AP Inter 2nd Year

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5h Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5h

I.

Question 1.
Differentiate (3x² – 9x + 5)⁹ w.r.t. x
Solution:
Let y = (3x² – 9x + 5)⁹
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 9(3x² – 9x + 5)⁸\(\frac{\mathrm{d}}{\mathrm{dx}}\)(3x² – 9x + 5) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (f(x))ⁿ = n(f(x))^n-1\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= 9(3x² – 9x + 5)⁸[3(2x) – 9(1) + 0]
= 9(3x² – 9x + 5)⁸(6x – 9) = 27(3x² – 9x + 5)⁸(2x – 3).

Question 2
Differentiate sin³ x + cos⁶ x w.r.t. x
Solution:
Let y = sin³ x +cos⁶ x = (sin x)³ + (cos x)⁶
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3(sin x)²\(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x + 6(cos x)⁵\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (f(x))ⁿ = n(f(x))^n-1\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= 3 sin² x cos x – 6 cos⁵ x sin x
= 3 sin x cos x(sin x – 2 cos⁴ x)

II.

Question 1.
Differentiate (5x)^{3 cos 2x} w.r.t. x
Solution:
Let y = (5x)^{3 cos 2x} ………….. (i)
Taking logs of both sides of (1) we have
logy = log(5x)^{3 cos 2x} = 3 cos 2x log(5x)
Differentiating both sides w.r.t. x, we have

Question 2.
Differentiate sin^-1(x\(\sqrt{\mathrm{x}}\)), 0 ≤ x ≤ 1 w.r.t. x
Solution:
Let y = sin^-1(x\(\sqrt{\mathrm{x}}\)) = sin^-1 (x^3/2 [∵ x\(\sqrt{\mathrm{x}}\) = x¹ . x^1/2 = x^1+1/2 = x^3/2]
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{\sqrt{1-\left(x^{3 / 2}\right)^2}} \frac{d}{d x} x^{3 / 2}\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\)sin^-1f(x) = \(\frac{1}{\sqrt{1-(f(x))^2}} \frac{d}{d x}\)f(x)]
= \(\frac{1}{\sqrt{1-x^3}} \frac{3}{2} x^{1 / 2}\)
= \(\frac{3 \sqrt{x}}{2 \sqrt{1-x^3}}\)
= \(\frac{3}{2} \sqrt{\frac{x}{1-x^3}}\)

Question 3.
Differentiate cos (a cos x + b sin x), for some constant a and b. w.r.t. x
Solution:
Let y = cos(a cos x + b sin x) for some constants a and b.
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin(a cos x + b sin x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(a cos x + b sin x) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos f(x) = -sin f(x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= -sin(a cos x + b sin x)[-a sin x + b cos x]
= -(-a sin x + b cos x)sin(a cos x + b sin x)
= (a sin x – b cos x)sin(a cos x + b sin x).

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) if y = 12(1 – cost), x = 10(t – sint), – \(\frac{\pi}{2}\) < t < \(\frac{\pi}{2}\)
Solution:
Given that y = 12(1 – cos t) and x = 10(t – sin t)
Differentiating both equations wr.t. t, we haye
\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 12\(\frac{\mathrm{d}}{\mathrm{dt}}\)(1 – cost) = 12(0 + sin t) = 12 sin t
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 10\(\frac{\mathrm{d}}{\mathrm{dt}}\)(t – sin t) = 10(1 – cos t)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{12 \sin t}{10(1-\cos t)}=\frac{6}{5} \cdot \frac{2 \sin \frac{1}{2} \cos \frac{1}{2}}{2 \sin ^2 \frac{t}{2}}=\frac{6}{5} \frac{\cos \frac{1}{2}}{\sin \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2} .\)

Question 5.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution:
Given that sin (A + B) = sinA cosB + cosA sinB
Assuming A and B are functions of x and differentiating both sides w.r.t x, we have
cos(A+B)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(A+B) = sin A \(\frac{\mathrm{d}}{\mathrm{dx}}\)(cos B) + cos B\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin A) + cos A\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin B) + sin B\(\frac{\mathrm{d}}{\mathrm{dx}}\)(cosA)
⇒ cos(A+B)\(\left(\frac{\mathrm{dA}}{\mathrm{dx}}+\frac{\mathrm{dB}}{\mathrm{dx}}\right)\) = -sin A sin B\(\frac{\mathrm{dA}}{\mathrm{dx}}\) + cos B cos A\(\frac{\mathrm{dA}}{\mathrm{dx}}\) + cos A cos B\(\frac{\mathrm{dB}}{\mathrm{dx}}\) – sin B sin A\(\frac{\mathrm{dA}}{\mathrm{dx}}\)
=(cos A cos B – sin A sin B) \(\frac{\mathrm{dB}}{\mathrm{dx}}\) +(cos A cos B – sin A sin B)\(\frac{\mathrm{dA}}{\mathrm{dx}}\)
= (cos A cos B – sin A sin B)\(\left(\frac{\mathrm{dB}}{\mathrm{dx}}+\frac{\mathrm{dA}}{\mathrm{dx}}\right)\)
Cancelling \(\left(\frac{\mathrm{dB}}{\mathrm{dx}}+\frac{\mathrm{dA}}{\mathrm{dx}}\right)\) both sides, we have cos(A + B)= cos A cos B – sin A sin B

Question 6.
If y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\), prove that \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Solution:
Given that y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Expanding the determinant along the first row,
y = f(x)(mc – nb) – g(x)(lc – na) + h(x)(lb – ma)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (mc – nb)\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x) – (lc – na)\(\frac{\mathrm{d}}{\mathrm{dx}}\)g(x) + (lb – ma)\(\frac{\mathrm{d}}{\mathrm{dx}}\)h(x)
= (mc – nb)f'(x) – (lc – na)g'(x) + (lb – ma)h'(x) ………… (i)
R.H.S = \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
= f'(x)(mc – nb) – g'(x)(lc – na) + h'(x)(lb – ma)
= (mc – nb)f'(x) – (lc – na)g'(x) + (lb – ma)h'(x) ………….. (ii)
From (i)and (ii), we have L.H.S. = RHS.

Question 7.
Differentiate the function (log x)^{cos x} w.r.t. x.
Solution:
Let y = (log x)^{cos x} ………………. (i)
⇒ log y = log(log x)^{cos x} = cos x log(logx) [∵ log mⁿ = n log m]
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)[cos x log(log x)]
⇒ \(\frac{1}{y}\frac{\mathrm{d}}{\mathrm{dx}}\) = cos x \(\frac{\mathrm{d}}{\mathrm{dx}}\) log (log x) + log(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) cos x [By Product rule]
= cos x \(\frac{1}{\log x} \frac{d}{d x}\)log x + log(log x)(- sin x) = \(\frac{\cos x}{\log x} \frac{1}{x}\) -sin x log(log x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y[\(\frac{\cos x}{x \log x}\) – sin x log(log x)]
Putting the value of y from (i), \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (log x)^{c0s x}[\(\frac{\cos x}{x \log x}\) – sin x log(log x)]

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of the function xy = e^{(x – y)}
Solution:
Given that xy = = e^{(x – y)} ⇒ log(xy) = log e^{(x – y)}
⇒ log x + log y = (x – y) log e ⇒ log x + logy = x – y (∵ log e = 1)
Differentiating both sides w.r.t. x, we have \(\frac{\mathrm{d}}{\mathrm{dx}}\)log x + \(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)x – \(\frac{\mathrm{d}}{\mathrm{dx}}\)y

Question 9.
Differentiate the function x (log x)^{log x}, x > 1 w.r.t x.
Solution:
Let y = (log x)^{log x}, x >1 …………… (i)
Taking log of both sides of (i), we have
log y = log(log x)^{log x} = log x log(log x)
Differentiating both sides w.r.t x,we have \(\frac{\mathrm{d}}{\mathrm{dx}}\)(logy) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (log x log(log x))

III.

Question 1.
Differentiate cot^-1\(\left[\frac{\sqrt{1+\sin \mathrm{x}}+\sqrt{1-\sin \mathrm{x}}}{\sqrt{1+\sin \mathrm{x}}-\sqrt{1-\sin \mathrm{x}}}\right]\), 0 < x < \(\frac{\pi}{2}\) w.r.t. x
Solution:

Question 2.
Differentiate (sin x – cos x)^{(sin x – cos x)}, \(\frac{\pi}{4}\) < x < \(\frac{3\pi}{4}\) w.r.t. x
Solution:
Let y = (sin x – cos x)^{(sin x – cos x)} …………… (i)
⇒ logy =log(sin x – cos x)^{(sin x – cos x)} = (sin x – cos x)log(sin x – cos x)
Differentiating both sides w.r.t x,we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\)log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x – cos x)log(sin x – cos x)
\(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = (sin x – cos x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)log(sin x – cos x) + log(sin x – cos x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x – cos x)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = (sin x – cos x)\(\frac{1}{(\sin x-\cos x)} \frac{d}{d x}\) (sin x – cos x) + log(sin x – cos x)(cos x + sin x)
=(cos x + sin x) + (cos x + sin x)log(sin x – cos x)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = (cos x + sin x)[1 + log(sin x – cos x)]
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y(cos x + sin x)[1 + log(sin x – cosx)]
Putting the value of y from (i),
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (sin x – cos x)^{(sin x – cos x)} (cos x + sin x)[1 + log(sin x – cos x)]

Question 3.
Differentiate x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0 w.r.t. x
Solution:
Let y = x^x + x^a + a^x + a^a
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) x^x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^a + \(\frac{\mathrm{d}}{\mathrm{dx}}\)a^x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)a^a
= \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^x + ax^a-1 + a^x log a + 0 [∵ a^a is constant as 3³ = 27 is constant]
= \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^x + ax^a-1 + a^x log a ………….. (i)
To find \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x^x) We take u = x^x ……………… (ii)
Taking log on both sides of eqn (ii),we have log u = log x^x = x log x
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log u = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x log x) ⇒ \(\frac{1}{u} \frac{d u}{d x}\) = x \(\frac{\mathrm{d}}{\mathrm{dx}}\)(log x) + log x\(\frac{\mathrm{d}}{\mathrm{dx}}\)x (Product Rule)
= x\(\frac{1}{\mathrm{x}}\) + log x.1 = 1 + log x ⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = u(1 + log x)
\(\frac{\mathrm{d}}{\mathrm{dx}}\) x^x = x^x(1 + log x) [By putting the value of u from (ii)]
Putting this value in eqn. (i), \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x^x(1 + log x) + ax^a-1 + a^x log a.

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\), if y = sin^-1x + sin^-1\(\sqrt{1-x^2}\), 0 < x < 1
Solution:

Question 5.
If x\(\sqrt{1+\mathrm{y}}\) + y\(\sqrt{1+\mathrm{x}}\) = 0, for -1 < x < 1, prove that \(\frac{d y}{d x}=-\frac{1}{(1+x)^2}\)
Solution:
Given that x\(\sqrt{1+\mathrm{y}}\) + y\(\sqrt{1+\mathrm{x}}\) = 0  …………. (i)
We shall first find y in terms of x
From eqn. (i), xx\(\sqrt{1+\mathrm{y}}\) = -y\(\sqrt{1+\mathrm{x}}\)
Squaring on both sides,
x²(1 + y) = y²(1 + x)
⇒ x² + x²y = y² + y²x or x² – y² = -x²y + y²x
⇒ (x – y)(x + y) = -xy(x – y)
Dividing both sides by (x – y) ≠ 0 (∵ x ≠ y)
x + y = -xy ⇒ y + xy = -x ⇒ y(1 + x) = -x
⇒ y = –\(\frac{x}{1+x}\)
Now differentiating both sides wrt.x, we have
\(\frac{d y}{d x}=-\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^2}=-\frac{(1+x) \cdot 1-x \cdot 1}{(1+x)^2}=-\frac{1}{(1+x)^2} .\)

Question 6.
If (x – a)² + (y – b)² = c², for some c > 0, prove that \(\frac{\left[1+{\frac{d y}{d x}^2}\right]^3}{d^2 y}\) is a constant independent of a and b.
Solution:
Given that (x – a)² + (y – b)² = c² …………. (i)
Differentiating both sides of eqn. (i) w.r.t. x,

Putting (x – a)² + (y – b)² = c² from (i)
= \(\frac{\left(c^2\right)^{3 / 2}}{-c^2}=\frac{-c^3}{c^2}\) = -c which is a constant and is independent of a and b

Question 7.
If cos y = x cos (a + y), with cos a ≠ ± , prove that \(\frac{d y}{d x}=\frac{\cos ^2(a+y)}{\sin a}\)
Solution:
Given that cos y = x cos (a + y)

Question 8.
If x = a(cos t + t sin t) and y = a(sin t – t cos t). find \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\).
Solution:
Given x = a(cos t + t sin t) and y = a(sin t – t cos t),
Differentiating both eqns. w.r.t. t,we have
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = a(-sin t + t\(\frac{\mathrm{d}}{\mathrm{dt}}\)sin t + sin t\(\frac{\mathrm{d}}{\mathrm{dt}}\)t = a(-sin t + t cos t + sin t) = at cos t
\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = a(cost – \(\frac{\mathrm{d}}{\mathrm{dt}}\)(t cos t)) = a(cos t – t\(\frac{\mathrm{d}}{\mathrm{dt}}\)cos t – cos t \(\frac{\mathrm{d}}{\mathrm{dt}}\)t) a(cos t + t sin t – cos t) = at sin t
∴ y = u + v
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\text { at } \sin t}{\text { at } \cos t}=\frac{\sin t}{\cos t}\) = tan t
Now differentiating both sides w.r.t.. x,we have \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (tan t) sec² t\(\frac{\mathrm{d}}{\mathrm{dt}}\)(t)
= sec² t\(\frac{\mathrm{d}}{\mathrm{dt}}\) = sec² t\(\left(\frac{1}{a t \cos t}\right)\) ………..(By(i))
= sec² t\(\left(\frac{\sec t}{\mathrm{at}}\right)=\frac{\sec ^3 t}{\mathrm{at}}\)

Question 9.
If y = e^{a cos-1x}, -1 ≤ x ≤ 1, show that (1 – x²)\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) – x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – a²y = 0
Solution:
Given that y = e^{a cos-1x}

Question 10.
Find the derivative of the function \(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\), -2 < x < 2 with respect to x.
Solution:
Let y = \(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\)
Applying the Quotient rule, we have

Question 11.
Find the derivative of the function x^x2-3 + (x – 3)^x2, for x > 3 with respect to x.
Solution:
Let y = x^x2-3 + (x – 3)^x2 for x > 3
Put u = x^x2-3 and v = (x – 3)^x2
∴ y = u + v
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) + \(\frac{\mathrm{dv}}{\mathrm{dx}}\) ………. (1)
Now u = x^(x2-3)
∴ Taking log of both sides, we have
log u = log x^(x2-3) = (x² – 3) log x.
Differentiating both sides w.r.t. x, we have
\(\frac{1}{\mathrm{u}}\frac{\mathrm{dy}}{\mathrm{dx}}\) = (x² – 3) \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x\(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 3) = (x² – 3)\(\frac{1}{\mathrm{x}}\) + log x(2x – 0)
⇒ \(\frac{1}{\mathrm{u}}\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^2-3}{x}\) + 2x log x
∴ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = u[\(\frac{x^2-3}{x}\) – 2 x log x]
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = x^{(x2 – 3)} \(\left(\frac{x^2-3}{x}+2 x \log x\right)\) [By putting u = x^x2-3]
Now consider v = (x – 3)^x2 ⇒ log v = log (x – 3)^x2 = x² log(x – 3)
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\)log v = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x² log(x – 3))
⇒ \(\frac{1}{\mathrm{v}}\frac{\mathrm{dv}}{\mathrm{dx}}\) = x²\(\frac{\mathrm{d}}{\mathrm{dx}}\) log(x – 3) + log(x – 3)\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²
= x²\(\frac{1}{x-3} \frac{d}{d x}\)(x – 3) + log(x – 3) . 2x
⇒ \(\frac{1}{\mathrm{v}}\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{x^2}{x-3}\) + 2x log(x – 3)
⇒ \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = v[\(\frac{x^2}{x-3}\) + 2x log(x – 3)]
= (x – 3)^x2[\(\frac{x^2}{x-3}\) + 2x log(x – 3)] ……………. (iii) [By putting v = (x – 3)^x2]
Putting values of \(\frac{\mathrm{du}}{\mathrm{dx}}\) and \(\frac{\mathrm{dv}}{\mathrm{dx}}\) from (ii) and (iii) in (i), we have
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x^(x2-3)[\(\frac{x^2-3}{x}\) + 2x log x] + (x – 3)^x2 [\(\frac{x^2}{x-3}\) + 2x log(x – 3)]

Question 12.
If f (x) = |x|³ show that f”(x) exists for all real x and find it.
Solution:
Given f (x) = |x|³ = x³ if x ≥ 0 ………… (i) [∵ |x| = x if x ≥ 0]
and f(x) = |x|³ = (-x)³ = -x³ if x < 0 (ii) [∵ |x| = -x if x < 0]
f'(x) = 3x² if x >0 and f'(x) = -3x² if x < 0 ……………. (iii) (At x = 0, we can’t write the value of f(x) by usual rule of derivatives because x = 0 is a partitioning point of values of f(x) given by (i) and (ii)) ∴ f'(x) = 6x if x > 0 and f'(x) = -6x if x < 0 …………… (iv) ∴ From (iv), f”(x) exists for all x > 0 and for all x < 0 i.e., for all x ∈ R except at x = 0
(i) Let us discuss derivability of f(x) at x = 0
L f'(0) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) \(\frac{f(x)-f(0)}{x-0}\) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) \(\frac{-x^3-0}{x}\) [By (ii) and (i)]
= \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) -x² = 0 (On putting x = 0)
R f'(0) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) \(\frac{f(x)-f(0)}{x-0}\) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) \(\frac{-x^3-0}{x-0}\) [By (i)]
= \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) x² = 0 (On putting x = 0)
∴ Lf'(0) = Rf'(0) = 0
∴ f(x) is derivable at x = 0 and f'(0) = 0

Referring to the AP Inter 2nd Year Maths Study Material Chapter 8 Application of Integrals Exercise 8b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Application of Integrals Solutions Exercise 8b

Question 1.
Find the area under the given curves and given lines:
(i) y = x², x = 1, x = 2 and x-axis
(ii) y = x⁴, x = 1, x = 5 and x-axis
Solution:
(i) y = x², x = 1, x = 2 and x-axis
Required Area A = \(\int_1^2 y d x=\int_1^2 x^2 d x=\left[\frac{x^3}{3}\right]_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3} \text { sq. units }\)

(ii) y = x⁴, x = 1, x = 5 and x-axis
Required Area A =\(\int_1^5 y d x=\int_1^5 x^4 d x=\left[\frac{x^5}{5}\right]_1^5=\frac{(5)^5}{5}-\frac{1}{5}=(5)^4-\frac{1}{5}=625-\frac{1}{5}=624.8 \text { sq. units }\)

Question 2.
Sketch the graph of y = |x + 3| and evaluate \(\int_{-6}^0|x+3| d x\).
Solution:

(x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0
∴ \(\int_{-6}^0|(x+3)| d x=-\int_{-6}^{-3}(x+3) d x+\int_{-3}^0(x+3) d x\)
= \(-\left[\frac{x^2}{2}+3 x\right]_{-6}^{-3}+\left[\frac{x^2}{2}+3 x\right]_{-3}^0\)
= \(-\left[\left(\frac{(-3)^2}{2}+3(-3)\right)-\left(\frac{(-6)^2}{2}+3(-6)\right)\right]+\left[0-\left(\frac{(-3)^2}{2}+3(-3)\right)\right]=-\left[-\frac{9}{2}\right]-\left[-\frac{9}{2}\right]\) = 9 sq. units

Question 3.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution:
Area bounded by the sine curve = Area OAB + Area BCD
Area (OAB) + Area(BCD) = \(\int_0^\pi \sin x d x+\left|\int_\pi^{2 \pi} \sin x d x\right|\)

= \([-\cos x]_0^\pi+\left|[-\cos x]_\pi^{2 \pi}\right|=[-\cos \pi+\cos 0]+|-\cos 2 \pi+\cos \pi|\)
= 1 + 1 + |(-1 – 1)| = 2 + |-2| = 2 + 2 = 4 sq. units

Question 4.
Find the area cut off between the line y = 0 and the parabola y = x² – 4x + 3
Solution:
Solving the given equations, we have x² – 4x + 3 = 0
⇒ (x – 1)(x – 3) = 0 ⇒ x = 1, 3
Also the given curve lies below the x-axis

Required area A = \(\int_1^3-y d x=-\int_1^3\left(x^2-4 x+3\right) d x=-\left[\frac{x^3}{3}-\frac{4 x^2}{2}+3 x\right]_1^3\)
= \(\left[9-18+9-\frac{1}{3}+2-3\right]=-\left[-\frac{1}{3}-1\right]=\frac{4}{3} \text { sq. units }\)

Question 5.
Find the area enclosed between the curve y = x², X-axis and the lines x = -1, x = 2
Solution:
The area bounded by the curve y = x², the x-axis and the lines x = -1, x = 2 is
A = \(\int_{-1}^2 \mathrm{ydx}=\int_{-1}^2 \mathrm{x}^2 \mathrm{dx}=\left[\frac{\mathrm{x}^3}{3}\right]_{-1}^2=\left(\frac{8}{3}\right)-\left(\frac{-1}{3}\right)=\frac{8}{3}+\frac{1}{3}=\frac{9}{3}\) = 3 sq. units

Question 6.
Find the area bounded between the curve y² = 2x + 1 and x = 0.
Solution:
Solving y² – 1 = 2x and x = 0, we get y² – 1 = 0 ⇒ y= ±1

The parabola y² – 1 = 2x meets the X-axis at y = 1 and y = -1.
As the curve is symmetric about the X-axis and lies left side to the Y-axis, the area bounded by the curve and the Y-axis is
A = \(2 \int_0^1(-x) d y=2 \int_0^1-\left(\frac{y^2-1}{2}\right) d y=\int_0^1-\left(y^2-1\right) d y=\int_0^1\left(1-y^2\right) d y=\left[y-\frac{y^3}{3}\right]_0^1=1-\frac{1}{3}=\frac{2}{3} \text { sq. units }\)

Question 7.
Find the area enclosed by the line y = 3x and curve y = 6x – x².
Solution:

The given curves are y = 3x …………(1); y = 6x – x² …..(2)
Solving (1), (2) we get 3x = 6x – x² ⇒ x² – 3x = 0 ⇒ x = 0,3
The upper boundary curve is y = 6x – x²
The lower boundary curve is y = 3x
∴ Required Area A = \(\int_0^3\left(\left(6 x-x^2\right)-3 x\right) d x=\int_0^3\left(3 x-x^2\right) d x=\left[\frac{3 x^2}{2}-\frac{x^3}{3}\right]_0^3=\frac{27}{2}-9=\frac{9}{2} \text { sq.n units }\)

Question 8.
Find the area between curve y = x³ + 3 and lines y = 0, x = -1, x = 2
Solution:
The area bounded by the curve y = x³ + 3, the x-axis and the lines x = -1, x = 2 is
A = \(\int_{-1}^2 y d x=\int_{-1}^2\left(x^3+3\right) d x=\left[\frac{x^4}{4}+3 x\right]_{-1}^2=\left[\left(\frac{16}{4}+6\right)-\left(\frac{1}{4}-3\right)\right]=\frac{40}{4}+\frac{11}{4}=\frac{51}{4} \text { sq. units }\)

Question 9.
Find the area between curve y² = 3x and line x = 3.
Solution:
Solving x = 3 and y² = 3x we have y² = 3(3) = 9 ⇒ y = ±3

∴ Required area A
\(=\int_{-3}^3\left[3-\frac{y^2}{3}\right] d y=2 \int_0^3\left[3-\frac{y^2}{3}\right] d y=2\left[3 y-\frac{y^3}{9}\right]_0^3\) = 2(9 – 3) = 12 sq. units.

Question 10.
Find the area between curve y = x² and the line y = 2x.
Solution:
The given curves are y = x² …. (1); y = 2x … (2)

Solving (1), (2) we have x² = 2x ⇒ x² – 2x = 0
⇒ x(x – 2) = 0 ⇒ x = 0, 2
The upper boundary curve is y = 2x,
the lower boundary curve is y = x².
∴ The area enclosed between the curves is
A = \(\int_0^2\left(2 x-x^2\right) d x=\left[2 \cdot \frac{x^2}{2}-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 8 Application of Integrals Exercise 8a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Application of Integrals Solutions Exercise 8a

I.

Question 1.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
Solution:
We have \(\frac{x^2}{16}+\frac{y^2}{9}=1 \Rightarrow \frac{y^2}{9}=1-\frac{x^2}{16} \Rightarrow y^2=9\left(\frac{16-x^2}{16}\right) \Rightarrow y=\frac{3}{4} \sqrt{16-x^2}\)
Ares of the ellipse = 4 × Area(OAB)

A = \(4 \int_0^4 \mathrm{ydx}=4\left(\frac{3}{4}\right) \int_0^4 \sqrt{16-\mathrm{x}^2} \mathrm{dx}\)
= \(3\left[\frac{x}{2} \sqrt{16-x^2}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_0^4=3\left[2 \sqrt{16-16}+8 \sin ^{-1}(1)-0-8 \sin ^{-1}(0)\right]\)
= 3\(\left[\frac{8 \pi}{2}\right]\) = 3[4π]= 12π
∴ Ares of the given ellipse = 12π sq.units

Question 2.
Find the area of the region bounded by the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Solution:
We have \(\frac{x^2}{4}+\frac{y^2}{9}=1 \Rightarrow \frac{y^2}{9}=1-\frac{x^2}{4} \Rightarrow y^2=9\left(\frac{4-x^2}{4}\right) \Rightarrow y=\frac{3}{2} \sqrt{4-x^2}\)
Area of the ellipse A = 4 × Area(OAB)

A = \(4 \int_0^2 \mathrm{ydx}=4\left(\frac{3}{2}\right) \int_0^2 \sqrt{4-\mathrm{x}^2} \mathrm{dx}\)
= \(\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_0^2=6\left[\frac{2 \pi}{2}\right]=6 \pi\)
∴ Ares of the given ellipse = 6π sq.units

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5g Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5g

I.

Question 1.
Find the second order derivative of x² + 3x + 2
Solution:
Let y = x² + 3x + 2
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x + 3 . 1 + 0 = 2x + 3.
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)\) = 2(1) + 0 = 2

Question 2.
Find the second order derivative of x²⁰.
Solution:
Let y = x²⁰
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 20x¹⁹
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 20 × 19x¹⁸ = 380x¹⁸

Question 3.
Find the second order derivative of x . cos x
Solution:
Let y = x cos x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x + c0s x \(\frac{\mathrm{d}}{\mathrm{dx}}\)x [By Product Rule]
= -x sin x + cos x .
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = –\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x sin x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x
= -[x\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin x + sin x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x)] – sin x
= -(x cos x + sin x) – sin x = -x cos x – sin x – sin x
= -x cos x – 2 sin x = -(x cos x + 2 sin x).

Question 4.
Find the second order derivative of log x
Solution:
Let y = log x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{x}\)
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{d}{d x}\left(\frac{1}{x}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^-1
= (-1)x^-2 = \(\frac{-1}{x^2}\)

Question 5.
Find the second order derivative of tan^-1 x
Solution:
Let y = tan^-1 x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{1+x^2}\)
Again differentiating w.r.t. x, we have
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{1+\mathrm{x}^2}\right)\) = \(\frac{\left(1+x^2\right) \frac{d}{d x}(1)-1 \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2}\)
= \(\frac{\left(1+x^2\right) 0-(2 x)}{\left(1+x^2\right)^2}\) = \(\frac{-2 x}{\left(1+x^2\right)^2}\)

II.

Question 1.
Find the second order derivative of x³ log x
Solution:
Let y = x³ log x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x³\(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x \(\frac{\mathrm{d}}{\mathrm{dx}}\) x³ [By Product Rule]
= x³\(\frac{1}{\mathrm{x}}\) + (log x)3x² = x² + 3x² log x
Again differentiating w.r.t. x, we have
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) x² + 3\(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² log x) = 2x + 3[x² \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²]
= 2x + 3(x² . \(\frac{1}{\mathrm{x}}\) + (log x)2x) = 2x + 3(x + 2x log x)
= 2x + 3x + 6x log x = 5x + 6x log x
= x(5 + 6 log x)

Question 2.
Find the second order derivative of e^x sin 5x
Solution:
Let y = e^x sin 5x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin 5x + sin 5x\(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x [By Product Rule]
= e^x cos5x\(\frac{\mathrm{d}}{\mathrm{dx}}\)5x + sin5xe^x = e^x cos5 x5 + e^x sin5x
= e^x (5 cos 5x + sin 5x)
Again differentiating w.r.t. x using product rule, we have
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = e^x \(\frac{\mathrm{d}}{\mathrm{dx}}\)(5 cos 5x + sin 5x) + (5 cos 5x + sin 5x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x
= e^x(5(-sin 5x)5 + (cos 5x)5) + (5 cos 5x + sin 5x)e^x
= e^x(-25 sin 5x + 5 cos 5x + 5 cos 5x + sin 5x)
= e^x(10 cos 5x – 24 sin 5x)
= 2e^x(5 cos 5x – 12 sin 5x).

Question 3.
Find the second order derivative of e^6x cos 3x
Solution:
Let y = e^6x cos 3x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^6x\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos 3x + cos 3x\(\frac{\mathrm{d}}{\mathrm{dx}}\)e^6x [By Product Rule]
= e^6x (-sin 3x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(3x) + cos 3x . e^6x\(\frac{\mathrm{d}}{\mathrm{dx}}\)6x
= -e^6x sin 3x . 3 + cos 3x e^6x . 6
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6e^6x cos 3x – 3e^6x sin 3x …………… (1)
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (6e^6x cos 3x – 3e^6x sin 3x) = 6\(\frac{\mathrm{d}}{\mathrm{dx}}\)(e^6x cos 3x) – 3\(\frac{\mathrm{d}}{\mathrm{dx}}\)(e^6x sin 3x)
= 6[6e^6x cos 3x – 3e^6x sin 3x] – 3[sin 3x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(e^6x) + e^6x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin 3x)] [using (1)1
= 36e^6x cos 3x – 18e^6x sin 3x – 3[sin 3xe^6x 6 + e^6x cos 3×3]
= 36e^6x cos 3x – 18e^6x sin 3x – 18e^6x sin 3x – 9e^6x cos3x
= 27 e^6x cos 3x – 36e^6x sin 3x = 9e^6x (3 cos 3x – 4 sin 3x)

Question 4.
Find the second order derivative of log (log x)
Solution:
Let y = log (log x)

Question 5.
Find the second order derivative of sin(log x)
Solution:
Let y = sin(log x)

Question 6.
If y = 5 cos x – 3 sin x, prove that \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) + y = 0
Solution:
Given that y = 5 cos x – 3 sin x ……………. (i)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -5 sin x – 3 cos x
Again differentiating w.r.t. x,
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = -y = -5 cos x + 3 sin x
= -(5 cos x – 3 sin x) = -y (By (i))
⇒ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = -y
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) + y = 0

Question 7.
If y = cos^-1x. Find \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) in terms of y alone.
Solution:
Given that y = cos^-1x ⇒ x = cos y …………. (i)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-1}{\sqrt{1-x^2}}=\frac{-1}{\sqrt{1-\cos ^2 y}}=\frac{-1}{\sqrt{\sin ^2 y}}=\frac{-1}{\sin y}\) = -cosec y [By (i)]
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -cosec y …………. (ii)
Again differentiating both sides w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = –\(\frac{\mathrm{d}}{\mathrm{dx}}\) (cosec y) = -[-cosec y cot y \(\frac{\mathrm{dy}}{\mathrm{dx}}\)]
= cosec y cot y(-cosec y) = -cosec²y cot y.

Question 8.
If y 3 cos (log x) + 4 sin (log x). show that x²y₂ + xy₁ + y = 0
Solution:
Given that y = 3cos(log x)+ 4sin(log x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (y₁)= -3 sin(logx)\(\frac{\mathrm{d}}{\mathrm{dx}}\)logx + 4cos(log x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) logx
⇒ y₁ =- 3 sin(1ogx) \(\frac{1}{x}\) +4cos(logx). \(\frac{1}{x}\)
⇒ xy₁ = -3 sin(log x) +4 cos(log x)
Again differentiating both sides wrt. x,
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (xy₁) = -3 cos(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)log x – 4sin(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)log x
x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y₁ + y₁\(\frac{\mathrm{d}}{\mathrm{dx}}\)x = – 3 cos(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) – 4 sin(log x)\(\frac{1}{\mathrm{dx}}\) [By Product Rule]
⇒ xy₂ + y₁ = –\(\frac{[3 \cos (\log x)+4 \sin (\log x)]}{x}\)
⇒ x(xy₂ + y₁) = -[3 cos(log x)+ 4 sin(log x)]
⇒ x²y₂ + xy₁ = -y ⇒ x²y₂ + xy₁ + y = 0

Question 9.
If y = Ae^mx + Be^nx, show that \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) – (m + n) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + mny = 0
Solution:
Given that y =Ae^mx + Be^nx …………. (i)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = Ae^mx\(\frac{\mathrm{d}}{\mathrm{dx}}\)(mx) + Be^nx\(\frac{\mathrm{d}}{\mathrm{dx}}\)(nx) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) f(x)]
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = Ame^mx + Bne^nx …………. (ii)
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = Ame^mxm + Bne^nx.n = Am²e^mx + Bn²e^nx ……………. (iii)
Putting values of y, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) from(i), (ii) and (iii) in
L.H.S. = \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) – (m + n)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + mny
= Am²e^mx + Bn²e^nx – (m + n)(Ame^mx +Bne^nx) + mn(Ae^mx + Be^nx)
= Am²e^mx + Bn²e^nx – Am²e^mx – Bmne^nx – Anme^mx – Bn²e^nx + Amne^mx + Bnme^nx = 0
= R.H.S.

Question 10.
If y = 500e^7x + 600e^-7x, show that \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 49y
Solution:
Given y = 500e^7x + 600e^-7x …………. (i)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 500e^7x (7) + 600e^-7x(-7) = 500(7)e^7x – 600(7)e^-7x
Now \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 500(7)e^7x (7) – 600(7)e^-7x (7)
= 500(49)e^7x + 600(49)e^-7x
= 49[500e^7x + 600e^-7x] = 49 y ………………. [By (I)]
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 49y.

Question 11.
If e^y (x + 1) = 1, show that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\)
Solution:
Given that e^y (x + 1) = 1 ⇒ e^y = \(\frac{1}{x+1}\)
Taking logs of both sides, log e^y = log \(\frac{1}{x+1}\)
⇒ y loge = log 1 – log(x + 1)
⇒ y = -log(x + 1) [∵ log e = 1 and log 1 = 0]

Question 12.
If y = (tan^-1x)², show that (x² + 1)²y₂ + 2x (x² + 1)y₁ = 2
Solution:
Given that y = (tan^-1x)²
⇒ y₁ = 2(tan^-1x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)tan^-1x [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (f(x))ⁿ = n(f(x))^n-1\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
⇒ y₁ = 2(tan^-1x)\(\frac{1}{1+x^2}\)
⇒ y₁ = \(\frac{2 \tan ^{-1} x}{1+x^2}\)
⇒ (1 + x²)y₁ = 2 tan^-1x
Again differentiating both sides w.r.t. x,
(1 + x²)\(\frac{\mathrm{d}}{\mathrm{dx}}\)y₁ + y₁\(\frac{\mathrm{d}}{\mathrm{dx}}\)(1 + x²) = 2 . \(\frac{1}{1+x^2}\)
⇒ (1 + x²)y₂ + y₁ . 2x = \(\frac{2}{1+x^2}\)
⇒ (x² + 1)²y₂ + 2x(1 + x²)y₁ = 2.

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7j Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7j

I.

Question 1.
Evaluate \(\int_0^4|x-1| d x\)
Solution:
The given integral is |x – 1|
(x – 1) ≤ 0 when 0 ≤ x ≤ 1 and (x – 1) ≥ 0 when 1 ≤ x ≤ 4
I = \(\int_0^1|x-1| d x+\int_1^4|x-1| d x\) (\(\int_a^b f(x) d x=\int_b^c f(x) d x+\int_c^b f(x) d x\))
I = \(\int_0^1-(x-1) d x+\int_1^4(x-1) d x\)
= \(\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^4=\left(1-\frac{1}{2}-0\right)+\left[(8-4)-\left(\frac{1}{2}-1\right)\right]=\frac{1}{2}+\frac{9}{2}=5\)

Question 2.
Evaluate \(\int_2^8|x-5| d x\)
Solution:
Let I = \(\int_2^8|x-5| d x\)
As (x – 5) ≤ 0 on [2, 5] and (x – 5) ≥ 0 on [5, 8]
I = \(\int_2^5-(x-5) d x+\int_5^8(x-5) d x\) (∵ \(\int_a^b f(x)=\int_a^c f(x)+\int_c^b f(x)\))
= \(\left[\frac{\mathrm{x}^2}{2}-5 \mathrm{x}\right]_2^5+\left[\frac{\mathrm{x}^2}{2}-5 \mathrm{x}\right]_5^8=-\left[\frac{25}{2}-25-2+10\right]+\left[32-40-\frac{25}{2}+25\right]\) = 9

Question 3.
Evaluate \(\int_{-5}^5|x+2| d x\)
Solution:
Let I = \(\int_{-5}^5|x+2| d x\)
As, (x + 2) ≤ 0 on [-5, -2] and (x + 2) ≥ 0 and [-2, 5]
∴ \(\int_{-5}^5|x+2| d x=\int_{-5}^{-2}-(x+2) d x+\int_{-2}^5(x+2) d x\)
I = \(-\left[\frac{x^2}{2}+2 x\right]_{-5}^{-2}+\left[\frac{x^2}{2}+2 x\right]_{-2}^5\)
= \(-\left[\frac{(-2)^2}{2}+2(-2)-\frac{(-5)^2}{2}-2(-5)\right]+\left[\frac{(5)^2}{2}+2(5)-\frac{(-2)^2}{2}-2(-2)\right]\)
= \(-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]=-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4=29\)

Question 4.
Evaluate \(\int_0^1 x(1-x)^n d x\)
Solution:
Let I = \(\int_0^1 x(1-x)^n d x\)
∴ I = \(\int_0^1(1-x)(1-(1-x))^n d x=\int_0^1(1-x)(x)^n d x=\int_0^1\left(x^n-x^{n+1}\right) d x\)
= \(\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1\) [∵ \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\)]
= \(\left[\frac{1}{n+1}-\frac{1}{n+2}\right]=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}\)

Question 5.
Evaluate \(\int_0^2 x \sqrt{2-x} d x\)
Solution:

Question 6.
Evaluate \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) ……..(1)
⇒ I = \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\)
⇒ I = \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) …..(2)
Adding (1) and (2), we get 2I = \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
⇒ 2I = \(\int_0^{\frac{\pi}{2}} 1 . d x \Rightarrow 2 I=[x]_0^{\frac{\pi}{2}} \Rightarrow 2 I=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}\)

Question 7.
Evaluate \(\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}\)
Solution:

Question 8.
Evaluate \(\int_0^{\pi / 2} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x} d x\)
Solution:
We know \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)
I = \(\int_0^{\pi / 2} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x} d x\) ……..(1) = \(\int_0^{\pi / 2} \frac{\cos ^5\left(\frac{\pi}{2}-x\right)}{\sin ^5\left(\frac{\pi}{2}-x\right)+\cos ^5\left(\frac{\pi}{2}-x\right)} d x\)
= \(\int_0^{\pi / 2} \frac{\sin ^5 x d x}{\sin ^5 x+\cos ^5 x}\) …………..(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \frac{\cos ^5 x+\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x=\int_0^{\pi / 2} 1 d x=[x]_0^{\pi / 2}=\frac{\pi}{2}-0=\frac{\pi}{2}\)
∴ 2I = \(\frac{\pi}{2}\) ⇒ I = \(\frac{\pi}{4}\)

Question 9.
Evaluate \(\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\) ………(1)
⇒ I = \(\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x\)
⇒ I = \(\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x} d x\)
Adding (1) and (2), we get 2I = \(\int_0^{\frac{\pi}{2}} \frac{0}{1+\sin x \cos x} d x\) ⇒ I = 0

Question 10.
Evaluate \(\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x\)
Solution:
Let I = \(\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x\) ……(1)
We know that, (\(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\))
I = \(\int \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x\) ……….(2)
Adding (1) and (2), we get 2I = \(\int_0^a \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x\)
⇒ 2I = \(\int_0^a 1 . d x \Rightarrow 2 I=[x]_0^a \Rightarrow 2 I=a \Rightarrow I=\frac{a}{2}\)

Question 11.
Evaluate \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x\) dx
Solution:
Let I = \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x\) dx ………….(1)
Now sin⁷(-x) = [sin(-x)]⁷ = (-sin x)⁷ = -sin x, ⇒ sin² x is an odd function
If f(x) is an odd function, then \(\int_{-a}^a f(x) d x=0\)
∴ I = \(\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x\) dx = 0

Question 12.
Evaluate \(\frac{\int_{-\pi}^{\frac{\pi}{2}}}{2} \sin ^2 x d x\)
Solution:
Let I = \(\frac{\int_{-\pi}^{\frac{\pi}{2}}}{2} \sin ^2 x d x\)
As sin²(-x) = [sin(-x)]² = (-sin x)² = sin² x ∴ sin² x is an even function
If f(x) is an even function, then \(\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=2 \int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)

Question 13.
Evaluate \(\int_{-1}^1 x^{17} \cos ^4 x d x\)
Solution:
Let I = \(\int_{-1}^1 x^{17} \cos ^4 x d x\)
Let f(x) = x¹⁷ cos⁴ x ⇒ f(-x) = (-x)¹⁷ cos⁴(-x) = -x¹⁷ cos⁴ x = -f(x)
f(x) is an odd function
We know that if f(x) is an odd function, then \(\int_{-a}^a f(x) d x=0\)
∴ I = \(\int_{-1}^1 x^{17} \cos ^4 x d x=0\)
Hence proved.

Question 14.
Evaluate \(\int_0^{2 \pi} \cos ^5 x d x\)
Solution:
Let I = \(\int_0^{2 \pi} \cos ^5 x d x\) ……….(1)
We have cos⁵(2π – x) = cos⁵ x
∴ I = \(2 \int_0^\pi \cos ^5 x d x\) ⇒ I = 2(0) = 0 [∵cos⁵(π – x) = – cos⁵x]

III.

Question 1.
Evaluate \(\int_0^{\pi / 4} \log (1+\tan x) d x\)
Solution:
We know \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)

Question 2.
Evaluate \(\int_0^1 \frac{\log (1+x)}{1+x^2} d x\)
Solution:
Put x = tan θ ⇒ dx = sec² θdθ. Also, x = 0 ⇒ θ = 0; x = 1 θ = \(\frac{\pi}{4}\)
Now, 1 + x² = 1 + tan² θ = sec² θ

Question 3.
Evaluate \(\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x\)
Solution:
Consider \(\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x\)
⇒ I = \(\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log (2 \sin x \cos x)) d x\)
[∵ sin2x = 2sin xcos x
log(abc) = log a + lob b + log c]
I = \(\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin x-\log \cos x-\log 2) d x\)
⇒ I = \(\int_0^{\frac{\pi}{2}}[\log \sin x-\log \cos x-\log 2] d x \ldots(1)\)
⇒ I = \(\left.\int_0^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right)-\log \cos \left(\frac{\pi}{2}-x\right)-\log 2\right] d x\)
⇒ I = \(\int_0^{\frac{\pi}{2}}[\log \cos x-\log \sin x-\log 2] d x\) ………(2)
Adding (1) and (2), we get
2I = \(\int_0^{\frac{\pi}{2}}(-\log 2-\log 2) \mathrm{dx} \Rightarrow 2 \mathrm{I}=-2 \log 2 \int_0^{\frac{\pi}{2}} 1 . \mathrm{dx}\)
⇒ I = \(-\log 2\left[\frac{\pi}{2}\right] \Rightarrow \mathrm{I}=\frac{\pi}{2}(-\log 2) \Rightarrow \mathrm{I}=\frac{\pi}{2}\left[\log \frac{1}{2}\right] \Rightarrow \mathrm{I}=\frac{\pi}{2} \log \frac{1}{2}=\frac{-\pi}{2} \log 2\)

Question 4.
Evaluate \(\int_0^\pi \frac{x}{1+\sin x} d x\)
Solution:

Question 5.
Find the integral of \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Solution:
Consider I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)

We know that if f(x) is an even function, then
I = \(2 \int_0^{\frac{\sqrt{3}-1}{2}} \frac{d t}{\sqrt{1-t^2}}=\left[2 \sin ^{-1} t\right]_0^{\frac{\sqrt{3}-1}{2}}=2 \sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\)

Question 6.
Evaluate \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\)
Solution:
Here, we take the substitution sinx – cosx = t. Then (cos x + sin x) = dt
Now x = 0 ⇒ t = sin0 – cos0 = 0 – 1 = -1 and x = π/4 ⇒ t = \(\)
Also, (sinx – cosx)² = t² ⇒ sin²x + cos²x – 2sinxcosx = t² ⇒ 1 – sin2x = t² ⇒ sin2x = 1 – t²
∴ 9 + 16sin2x = 9 + 16(1 – t²) = 9 + 16 – 16t² = 25 – 16t²
∴ I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x=\int_{-1}^0 \frac{d t}{25-16 t^2}=\int_{-1}^0 \frac{d t}{5^2-(4 t)^2}=\frac{1}{4} \cdot \frac{1}{2(5)} \cdot \log \left[\frac{5+4 t}{5-4 t}\right]_{-1}^0\)
= \(\frac{1}{40}\left[\log \left[\frac{5+0}{5-0}\right]-\log \left[\frac{5-4}{5+4}\right]\right]=\frac{1}{40}\left[\log 1-\log \frac{1}{9}\right]\)
= \(\frac{1}{40}\left[0-\log 9^{-1}\right]=\frac{1}{40}[\log 9]=\frac{\log 3^2}{40}=\frac{2 \log 3}{40}=\frac{\log 3}{20}\)

Question 7.
Find the integral of \(\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x\)
Solution:

Put tan²x = t ⇒ 2 tan x sec² xdx = dt
When x = 0 we have t = 0 and when x = \(\frac{\pi}{4}\), t = 1
∴ I = \(\frac{1}{2} \int_0^1 \frac{\mathrm{dt}}{1+\mathrm{t}^2}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_0^1=\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]=\frac{1}{2}\left[\frac{\pi}{4}\right]=\frac{\pi}{8}\)

Question 8.
Find the integral of \(\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x\)
Solution:

= \(-\frac{\pi}{6}+\frac{2}{3} \int_0^{\frac{\pi}{2}} \frac{2 \sec ^2 x}{1+4 \tan ^2 x} d x\) ……….(1)
Consider, \(\int_0^{\frac{\pi}{2}} \frac{2 \sec ^2 x}{1+4 \tan ^2 x} d x\)
Put, 2 tan x = t ⇒ 2 sec² xdx = dt
When x = 0 we have t = 0 and x = \(\frac{\pi}{2}\) we have t = ∞
∴ \(\int_0^{\frac{\pi}{2}} \frac{2 \sec ^2 x}{1+4 \tan ^2 x} d x=\int_0^{\infty} \frac{d t}{1+t^2}=\left[\tan ^{-1} t\right]_0^{\infty}=\left[\tan ^{-1}(\infty)-\tan ^{-1}(0)\right]=\frac{\pi}{2}\)
∴ from (1), we get I = \(-\frac{\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}\right]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)

Question 9.
Find the integral of \(\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x\)
Solution:

Question 10.
Find the integral of \(\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x=\int_0^{\frac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x\)
Put sin x = t ⇒ cos xdx= dt
When x = 0 we get t = 0 and when x = \(\frac{\pi}{2}\) we get t = 1 ⇒ I = \(2 \int_0^1 \mathrm{t} \tan ^{-1}(\mathrm{t}) \mathrm{dt}\) …………..(1)
Now \(\int t \cdot \tan ^{-1} t d t=\tan ^{-1} t \int t d t-\int\left[\frac{d}{d t}\left(\tan ^{-1} t\right) \int t d t\right] d t\)

Question 11.
Find the integral of \(\int_1^4[|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|] \mathrm{dx}\)
Solution:
Let I = \(\int_1^4[|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|] \mathrm{dx} \Rightarrow \mathrm{I}=\int_1^4|\mathrm{x}-1| \mathrm{dx}+\int_1^4|\mathrm{x}-2| \mathrm{dx}+\int_1^4|\mathrm{x}-3| \mathrm{dx}\)
I = I₁ + I₂ + I₃ ………(A)
Where, I₁ = \(\int_1^4|x-1| d x, I_2=\int_1^4|x-2| d x \text { and } I_3=\int_1^4|x-3| d x\)
I₁ = \(\int_1^4|x-1| d x\)
(x – 1) ≥ 0 for 1 ≤ x ≤ 4
∴ I₁ =\(\int_1^4(x-1) d x \Rightarrow I_1=\left[\frac{x^2}{2}-x\right]_1^4=\left[8-4-\frac{1}{2}+1\right]=\frac{9}{2}\) ………………(1)
I₂ = \(\int_1^4|x-2| d x\)
x – 2 ≤ 0 for 1 ≤ x ≤ 2 and x – 2 ≥ 0 for 2 ≤ x ≤ 4
∴ I₂ = \(\int_1^2(2-x) d x+\int_2^4(x-2) d x \Rightarrow I_2=\left[2 x-\frac{x^2}{2}\right]_1^2+\left[\frac{x^2}{2}-2 x\right]_2^4\)
⇒ I₂ = \(\left[4-2-2+\frac{1}{2}\right]+[8-8-2+4] \Rightarrow I_2=\frac{1}{2}+2=\frac{5}{2}\) ………….(2)
⇒ I₃ = \(\int_1^4|x-3| d x\)
x – 3 ≤ 0 for 1 ≤ x ≤ 3 and x – 3 ≥ 0 for 3 ≤ x ≤ 4
∴ I₃ = \(\int_1^3(3-x) d x+\int_3^4(x-3) d x\)
⇒ I₃ = \(\left[3 x-\frac{x^2}{2}\right]_1^3+\left[\frac{x^2}{2}-3 x\right]_3^4 \Rightarrow I_3=\left[9-\frac{9}{2}-3+\frac{1}{2}\right]+\left[8-12-\frac{9}{2}+9\right]\)
⇒ I₃ = \([6-4]+\left[\frac{1}{2}\right]=\frac{5}{2}\) ………….(3)
From equations (1), (2), (3) and (A), we get I = \(\frac{9}{2}+\frac{5}{2}+\frac{5}{2}=\frac{19}{2}\)

Question 12.
Evaluate \(\int_0^\pi \log (1+\cos x) d x\)
Solution:
Consider I = \(\int_0^\pi \log (1+\cos x) d x\) ………….(1) (∵ \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\))
⇒ I = \(\int_0^\pi \log [1+\cos (\pi-x)] d x \Rightarrow I=\int_0^\pi \log (1-\cos x) d x\) ……(2)
Adding (1) and (2), we get 2I = \(\int_0^\pi[\log (1+\cos x)+\log (1-\cos x)] d x\)
2I = \(\int_0^\pi \log \left(1-\cos ^2 x\right) d x \Rightarrow 2 I=\int_0^\pi \log \left(\sin ^2 x\right) d x\) [∵ log xⁿ = n logx]
⇒ 2I = \(2 \int_0^\pi \log (\sin x) d x \Rightarrow I=\int_0^\pi \log (\sin x) d x\) ……..(3)
∴ sin(π – x) = sin x
We know that \(\int_0^{2 \mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=2 \int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx} \text { if } \mathrm{f}(2 \mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})\)
∴ I = 2\(2 \int_0^{\frac{\pi}{2}} \log \sin x d x \quad \ldots(4) \Rightarrow I=2 \int_0^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x=2 \int_0^{\frac{\pi}{2}} \log \cos x d x\) …………..(5)
Adding (4) and (5), we get

Question 13.
Evaluate \(\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x\)
Solution:
\(\int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x\) Put 2x = t ⇒ 2dx = dt
When x = 1 we have t = 2 and when x = 2 we have t = 4
∴ \(\int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x=\frac{1}{2} \int_2^4 e^t\left(\frac{2}{t}-\frac{2}{t^2}\right) d t=\int_2^4 e^t\left(\frac{1}{t}-\frac{1}{t^2}\right) d t\)
= \(\left[e^t \cdot \frac{1}{t}\right]_2^4=\left[\frac{e^t}{t}\right]_2^4=\frac{e^4}{4}-\frac{e^2}{2}=\frac{e^2\left(e^2-2\right)}{4}\)

Question 14.
If f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4, then show that \(\)
Solution:
Let I = \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx} \ldots \ldots(1) \Rightarrow \int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{g}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\)
⇒ \(\int_0^a f(x) g(a-x) d x\) ………..(2)
Adding (1) and (2), we get 2I = \(\int_0^a\{f(x) g(x)+f(x) g(a-x)\} d x\)
⇒ 2I = \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x})\{\mathrm{g}(\mathrm{x})+\mathrm{g}(\mathrm{a}-\mathrm{x})\} \mathrm{dx}\)
⇒ 2I = \(\int_0^a f(x)(4) d x \quad[g(x)+g(a-x)=4] \Rightarrow I=2 \int_0^a f(x) d x\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5f Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5f

I. If x and y are connected parametrically by the equations without eliminating the parameter. find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = 2at², = at⁴
Solution:
Given that x = 2at² and y = at⁴.
Differentiating both eqns. w.r.t t we have
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(2at²) = 2a\(\frac{\mathrm{d}}{\mathrm{dt}}\)t² = 2a.2t = 4at
and \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(at⁴) = a\(\frac{\mathrm{d}}{\mathrm{dt}}\)t⁴ = a.4t³ = 4at³
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{4 a^3}{4 a t}\) = t².

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a cos θ, y = b cos θ
Solution:
Given that x = a cos θ and y = b cos θ.
Differentiating both eqns. w.r.t. θ have
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) (a cos θ) = a\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)cos θ = -a sin θ
and \(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(b cos θ) = b\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)cos θ = – b sin θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = sin t, y = cos 2t
Solution:
Given that x = sin t, y = cos 2t.
Differentiating both eqns. w.r.t t we have
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = cos t and \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = -sin 2t \(\frac{\mathrm{d}}{\mathrm{dt}}\) (2t) = -2 sin 2t
∴ \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = \(\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=-\frac{-2 \sin 2 \mathrm{t}}{\cos \mathrm{t}}=-2 \frac{2 \sin \mathrm{t} \cos \mathrm{t}}{\cos \mathrm{t}}\) = -4 sin t

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = 4t, y = \(\frac{4}{t}\).
Solution:
Given that x = 4t and y = \(\frac{4}{t}\).
Differentiating both eqns. w.r.t. x we have
\(\frac{d x}{d t}=\frac{d}{d t}(4 t)=4 \frac{d}{d t} t=4(1)=4 \text { and } \frac{d y}{d t}=\frac{d}{d t}\left(\frac{4}{t}\right)=\frac{t \frac{d}{d t}(4)-4 \frac{d}{d t} t}{t^2}=\frac{t(0)-4(1)}{t^2}=-\frac{4}{t^2}\)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\left(-\frac{4}{t^2}\right)}{4}=\frac{-1}{t^2}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:
Given that x = cos θ – cos 2θ and y = sin θ – sin 2θ
∴ \(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) (cos θ) – \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) cos 2θ = -sin θ – (-sin 2θ)\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)2θ = -sin θ + (sin 2θ)2 = 2sin 2θ – sin θ
and \(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = cos θ \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)sin 2θ = cos θ – cos 2θ \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) 2θ = cos θ – cos 2θ(2) = cos θ – 2 cos 2θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a (θ – sin θ). y = a(1 + cos θ)
Solution:
x = a(θ – sin θ) and y = a(1 + cos θ).
Differentiating both eqns. w.r.t. θ we have
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = a\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(θ – sin θ) = a[\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)θ – \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) sin θ] = a(1 – cosθ)
\(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = a\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(1 + cos θ) = a[\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(1) + \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(cos θ)] = a(0 – sin θ) = -a sin θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{-a \sin \theta}{a(1-\cos \theta)}=-\frac{\sin \theta}{1-\cos \theta}=-\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}}=-\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2} .\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = \(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\), y = \(\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Solution:
Given that x = \(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\) and y = \(\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Differentiating both eqns. w.r.t. x, we have

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a(cos t + log tan \(\frac{t}{2}\)), y = a sin t
Solution:

Question 9.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a sec θ, y = b tan θ
Solution:
Given that x = a sec θ and y = b tan θ.
Differentiating both eqns. w.r.t. θ we have
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = a sec θ tan θ and \(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = b sec² θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
=\(\frac{b \sec ^2 \theta}{a \sec \theta \tan \theta}=\frac{b \sec \theta}{a \tan \theta}\)
=\(\frac{b \cdot \frac{1}{\cos \theta}}{a \cdot \frac{\sin \theta}{\cos \theta}}=\frac{b}{\cos \theta} \cdot \frac{\cos \theta}{a \sin \theta}=\frac{b}{a \sin \theta}\)
=\(\frac{b}{a} \cos \sec \theta\)

Question 10.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a(cos θ + θ sin θ), y = a(sin θ – θ cos θ).
Solution:
Given that x = a(cos θ + θ sin θ), y = a(sin θ – θ cos θ).
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = a(-sin θ + θ cos θ + sin θ) = a θ cos θ
\(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = a[cos θ – (θ(-sin θ) + cos θ)] = a(cos θ + θ sin θ – cos θ] = a θ sin θ
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta\)

Question 11.
If x = \(\sqrt{a^{\sin ^{-1} t}}\), y = \(\sqrt{a^{\cos ^{-1} t}}\) show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7i Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7i

I.

Question 1.
Evaluate \(\int_0^1 \frac{x}{x^2+1} d x\)
Solution:
Put x² + 1 = t ⇒ 2xdx = dt
When x = 0 we have t = 1 and when x = 1 we have t = 2
∴ \(\int_0^1 \frac{\mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}=\frac{1}{2} \int_1^2 \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{2}\left[\left.\log |\mathrm{t}|\right|_1 ^2=\frac{1}{2}[\log 2-\log 1]=\frac{1}{2} \log 2\right.\)

Question 2.
Evaluate \(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x\)
Solution:
Put cosx = t ⇒ -sin xdx = dt
When x = 0 we have t = 1 and when x = \(\frac{\pi}{2}\) we have t = 0
⇒ \(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x=-\int_1^0 \frac{d t}{1+t^2}=-\left[\tan ^{-1} t\right]_1^0=-\left[\tan ^{-1} 0-\tan ^{-1} 1\right]=-\left[-\frac{\pi}{4}\right]=\frac{\pi}{4}\)

Question 3.
Evaluate \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\)
Solution:
\(\int_{-1}^1 \frac{d x}{x^2+2 x+5}=\int_{-1}^1 \frac{d x}{\left(x^2+2 x+1\right)+4}=\int_{-1}^1 \frac{d x}{(x+1)^2+2^2}\)
Put x + 1 = t ⇒ dx = dt
When x = -1 we have t = 0 and when x = 1 we have t = 2
\(\int_{-1}^1 \frac{\mathrm{dx}}{(\mathrm{x}+1)^2+2^2}=\int_0^2 \frac{\mathrm{dt}}{\mathrm{t}^2+2^2}=\left[\frac{1}{2} \tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^2=\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0=\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{8}\)

Question 4.
Evaluate \(\int_0^{\frac{\pi}{4}} 2 \tan ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} 2 \tan ^3 x d x\)

Question 5.
Evaluate \(\int_0^1 x e^x d x\)
Solution:
Let I = \(\int_0^1 x e^x d x\)
Let u = x and v = e^x and integrated by parts, we have
I = \(\left[x e^x\right]_0^1-\int_0^1\left[\left(\frac{d}{d x}(x)\right) \int e^x d x\right] d x=\left[x e^x\right]_0^1-\int_0^1 e^x d x=\left[x e^x\right]_0^1-\left[e^x\right]_0^1=\hat{e}-\hat{e}+1=1\)

Question 6.
Evaluate \(\int_1^2 \log x d x\)
Solution:
Let u = log x and v = 1, Now integrating by parts, we have
\(\int_1^2 \log x(1) d x=[\log x \cdot(x)]_1^2-\int_1^2(x) \frac{1}{x} d x\)
= [2 log2 – 1 log(1)] – \(\int_1^2 d x=2 \log 2-[x]_1^2=2 \log 2-[2-1]=2 \log 2-1\)

Question 7.
Evaluate \(\int_0^4 \frac{x^2}{1+x} d x\)
Solution:

II.

Question 1.
Evaluate \(\int_0^2 x \sqrt{x+2} d x\) (Put x + 2 = t²)
Solution:
\(\int_0^2 x \sqrt{x+2} d x\) Put x + 2 = t² ⇒ dx = 2tdt
When x = 0 we have t = \(\sqrt{2}\) and when x = 2 we have t = 2

Question 2.
Evaluate \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^4 \phi \cos \phi d \phi\)
Put sin Φ = t ⇒ cosΦ dΦ = dt
When Φ = 0 we have t = 0 and when Φ = \(\frac{\pi}{2}\) we have t = 1
∴ I = \(\int_0^1 \sqrt{t}\left(1-t^2\right)^2 d t=\int_0^1 t^{\frac{1}{2}}\left(1+t^4-2 t^2\right) d t=\int_0^1\left[t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right] d t\)
= \(\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}\right]_0^1=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}=\frac{154+42-132}{231}=\frac{64}{231}\)

Question 3.
Evaluate \(\int_0^2 \frac{d x}{x+4-x^2}\)
Solution:
\(\int_0^2 \frac{\mathrm{dx}}{\mathrm{x}+4-\mathrm{x}^2}=\int_0^2 \frac{\mathrm{dx}}{-\left(\mathrm{x}^2-\mathrm{x}-4\right)}=\int_0^2 \frac{\mathrm{dx}}{-\left[\mathrm{x}^2-2 \cdot \mathrm{x} \frac{1}{2}+\frac{1}{4}-\frac{1}{4}-4\right]}\)
= \(\int_0^2 \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^2-\frac{17}{4}\right]}=\int_0^2 \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}\)
Let x – \(\frac{1}{2}\) = t dx = dt when x = 0, t = \(-\frac{1}{2}\) and when x = 2, t = \(\frac{3}{2}\)

Question 4.
Evaluate \(\int_0^1 \sin ^{-1} x d x\)
Solution:
Let I = \(\int_0^1 \sin ^{-1} x d x \Rightarrow I=\int_0^1 \sin ^{-1} x \cdot 1 \cdot d x\)
Let u = sin^-1 x, v = 1 and integrated by parts, we get
I = \(\left[\sin ^{-1} x . x\right]_0^1-\int_0^1 \frac{1}{\sqrt{1-x^2}} x d x=\left[x \sin ^{-1} x\right]_0^1+\frac{1}{2} \int_0^1 \frac{(-2 x)}{\sqrt{1-x^2}} d x\)
Put 1 – x² = t ⇒ -2x dx = dt
When x = 0, t = 1 and when x = 1, t = 0
I = \(\left[x \sin ^{-1} x\right]_0^1+\frac{1}{2} \int_1^0 \frac{d t}{\sqrt{t}}=\left[x \sin ^{-1} x\right]_0^1+\frac{1}{2}[2 \sqrt{t}]_1^0=\sin ^{-1}(1)+[-\sqrt{1}]=\frac{\pi}{2}-1\)

Question 5.
Evaluate \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Solution:
Let I = \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Put x = tan θ ⇒ dx = sec² θdθ
When x = 0 we have θ = 0 and when x = 1 we have θ = \(\frac{\pi}{4}\)
∴ I = \(\int_0^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \sec ^2 \theta \mathrm{~d} \theta=\int_0^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^2 \theta \mathrm{~d} \theta\)
= \(\int_0^{\frac{\pi}{4}} 2 \theta \sec ^2 \theta \mathrm{~d} \theta=2 \int_0^{\frac{\pi}{4}} \theta \sec ^2 \theta \mathrm{~d} \theta\)
Taking u = θ and v = sec² θ and integrating by parts, we get

Question 6.
Evaluate \(\int_0^1 x \tan ^{-1} x d x\)
Solution:
Applying the “By Parts Rule:, we have \(\int_0^1 \tan ^{-1} x(x) d x=\left[\tan ^{-1} x\left(\frac{x^2}{2}\right)\right]_0^1-\int_0^1\left(\frac{x^2}{2}\right) \frac{1}{1+x^2} d x\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5e Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5e

Question 1.
Differentiate cos x . cos 2x . cos 3x w.r.t. x
Solution:
Let y = cos x cos 2x cos 3x ………… (i)
⇒ log(y) = log(cos x cos 2x cos 3x) = log(cos x) + log(cos 2x) + log(cos 3x)
Differentiating both sides w.r.t. x. we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\)log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)logcosx + \(\frac{\mathrm{d}}{\mathrm{dx}}\)logcos2x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)logcos3x.
∴ \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \frac{d}{d x} \cos x+\frac{1}{\cos 2 x} \frac{d}{d x} \cos 2 x+\frac{1}{\cos 3 x} \frac{d}{d x} \cos 3 x\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) logf(x) = \(\frac{1}{f(x)} \frac{d}{d x}\) f(x)]
= \(\frac{1}{\cos x}\)(-sin x) + \(\frac{1}{\cos 2x}\)(-sin 2x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{1}{\cos 3x}\)(-sin 3x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)3x
= -tan x – (tan2x)2 – tan3x(3)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -y(tanx + 2tan2x + 3tan3x)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -cosx cos2x cos3x(tan x + 2tan 2x +3tan3x). [∵ By putting the value of y from (i)]

Question 2.
Differentiate \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\) w.r.t. x.
Solution:
Let y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}=\left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]^{1 / 2}\) ……………. (i)
Taking log on both sides and simplifying we have
log y = \(\frac{1}{2}\)[log(x – 1) + log(x – 2) – log(x – 3) – log(x – 4) – log(x – 5)]
Differentiating both sides w.r.t. x we have

Question 3.
Differentiate x^x – 2^{sin x} w.r.t. x.
Solution:
Let y = x^x – 2^{sin x}
Put u = x^x and v = 2^{sin x} ∴ y = u – v
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) – \(\frac{\mathrm{dv}}{\mathrm{dx}}\) ………….. (i)
Now u = x^x
⇒ log u = log x^x = x log x [∵ log mⁿ = n log m]
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(log u) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x log x)
⇒ \(\frac{1}{u} \frac{d u}{d x}\) = x \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x \(\frac{\mathrm{d}}{\mathrm{dx}}\) x = x \(\frac{1}{x}\) + log x1
= 1 + log x
∴ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = u(1 + log x) = x^x(1 + log x) …………. (ii)
Again v = 2^{sin x}
∴ \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) 2^{sin x} = 2^{sin x} log2\(\frac{\mathrm{d}}{\mathrm{dx}}\)sinx [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) a^f(x) = a^f(x) log a\(\frac{d}{d x}\) f(x)]
⇒ \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 2^{sin x} (log 2)cos x = cos x . 2^{sin x} log 2 ………….. (iii)
Putting values from (ii) and (iii) in (i), we have
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x^x(1 + log x) – cos x 2^{sin x} log 2

Question 4.
Differentiate (x + 3)² . (x + 4)³ . (x + 5)⁴ w.r.t. x.
Solution:
Let y = (x + 3)² . (x + 4)³ . (x + 5)⁴ ………. (i)
log y = 2 log (x + 3) + 3 log(x + 4) + 4 log(x + 5)
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = 2\(\frac{\mathrm{d}}{\mathrm{dx}}\) log(x + 3) + 3\(\frac{\mathrm{d}}{\mathrm{dx}}\)log(x + 4) + 4\(\frac{\mathrm{d}}{\mathrm{dx}}\) log(x + 5)

Question 5.
Differentiate \(\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}\) w.r.t. x
Solution:

Question 6.
Differentiate (log x)^x + x^{log x} w.r.t. x.
Solution:
Let y = (log x)^x + x^{log x}
Put u = (log x)^x and v = x^{log x}
Then y = u + v ⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……….. (i)
Now u = (log x)^x
⇒ log u = log(log x)^x = x log(log x) [∵ log mⁿ = n log m]
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log u = \(\frac{\mathrm{d}}{\mathrm{dx}}\) [x log(log x)]
∴ \(\frac{1}{\mathrm{u}}\frac{\mathrm{d}}{\mathrm{dx}}\) = x \(\frac{\mathrm{d}}{\mathrm{dx}}\) log(log x) + log(log x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) x [By Product rule]

Question 7.
Differentiate (sin x)^x + sin^-1\(\sqrt{x}\) w.r.t. x.
Solution:
Let y = (sin x)^x + sin^-1\(\sqrt{x}\)
Put u = (sin x)^x and v = sin^-1\(\sqrt{x}\)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……….. (i)
Now u = (sin x)^x
∴ log u = log (sin x)^x = x log sin x

Question 8.
Differentiate x^{sin x} + (sin x)^{cos x} w.r.t. x.
Solution:
Let y = u + v
Put u = x^{sin x} and v = (sin x)^{cos x}
Then y = u + v
⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……….. (i)
Now u = x^{sin x}
∴ log u = log x^{sin x} = sin x log x
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log u = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x log x)
⇒ \(\frac{1}{\mathrm{u}}\frac{\mathrm{du}}{\mathrm{dx}}\) = sin x \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x
= sin x \(\frac{1}{\mathrm{x}}\) + (log x) cos x = \(\frac{\sin x}{x}\) + cos x log x
\(\frac{d u}{d x}=u\left(\frac{\sin x}{x}+\cos x \log x\right)=x^{\sin x}\left(\frac{\sin x}{x}+\cos x \log x\right)\) …………… (ii)
Again v =(sin x)^{c0s x} ⇒ log y = log(sin x)^{cos x} = cos x log sin x
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (log v) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)[cos x log sin x]
⇒ \(\frac{1}{\mathrm{v}}\frac{\mathrm{du}}{\mathrm{dx}}\) = cos x \(\frac{\mathrm{d}}{\mathrm{dx}}\) log sin x + log sin x\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x
= cos x \(\frac{1}{\sin x} \frac{d}{d x}\)(sin x) + log sin x(-sin x) = cot x . cos x – sin x log sin x
∴ \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = v(cos x cot x – sin x log sin x)
= (sin x)^{cos x}(cos x cot x – sin x log sin x) ……………. (iii)
Putting values of \(\frac{\mathrm{du}}{\mathrm{dx}}\) and \(\frac{\mathrm{dv}}{\mathrm{dx}}\) from (ii)and (in)in (i),we have
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x^{sin x}(\(\frac{\sin x}{x}\) + cos x log x) + (sin x)^{cos x}(cos x cot x – sin x log sin x)

Question 9.
Differentiate x^{x cos x} + \(\frac{x^2+1}{x^2-1}\) w.r.t. x.
Solution:
let y = x^{x cos x} + \(\frac{x^2+1}{x^2-1}\)
Putting x^{x cos x} = u and \(\frac{x^2+1}{x^2-1}\) = v
We have y = u + v ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) + \(\frac{\mathrm{dv}}{\mathrm{dx}}\) ……….. (i)
Now u = x^{x cos x}
Taking logarithms, log u = log x^{x cos x} = x cos x log x
Differentiating w.r.t. x, we have
\(\frac{1}{\mathrm{u}}\frac{\mathrm{du}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x cos x log x) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x) cos x log x + x \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos x) log x + x cos x \(\frac{\mathrm{d}}{\mathrm{dx}}\) (log x)
[∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uvw) + \(\frac{\mathrm{du}}{\mathrm{dx}}\)vw + u\(\frac{\mathrm{dv}}{\mathrm{dx}}\).w + uv\(\frac{\mathrm{dw}}{\mathrm{dx}}\)]
= 1 cos x log x + x(-sin x)log x + x cos x.\(\frac{1}{\mathrm{x}}\)
⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = u[cos x log x – x sin x log x + cos x]
= x^{x cos x}[cos x log x – x sin x log x + cos x] …………. (ii)

Question 10.
Differentiate (x cos x)^x + (x sin x)^1/x w.r.t. x.
Solution:
Let y = (x cos x)^x + (x sin x)^1/x
Putting (x cos x)^x = u and (x sin x)^1/x = v,
we have y = u + v
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) + \(\frac{\mathrm{dv}}{\mathrm{dx}}\) …………. (i)
Now u = (x cos x)^x
Taking logarithms,
log u = log(x cos x)^x = x log (x cos x) = x(log x + log cos x)
Differentiating w.r.t. x, we have
\(\frac{1}{u} \cdot \frac{d u}{d x}=x\left[\frac{1}{x}+\frac{1}{\cos x} \cdot(-\sin x)\right]\) + (log x + log cos x) . 1
⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\)[1 – x tan x + log(x cos x)] [∵ log a + log b = log ab]
= (x cos x)^x [1 – x tan x + log(x cos x)] ……………. (ii)
Also v = (x sin x)^1/x
Taking logarithms,
log v = log (x sin x)^1/x = \(\frac{1}{x}\) log (x sin x) = \(\frac{1}{x}\) (log x + log sin x)
Differentiating w.r.t. x, we have

Question 11.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of the function x^y + y^x = 1.
Solution:
Given that x^y + y^x = 1 ⇒ u + v = 1 where u = x^y and v = y^x
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (u) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (v) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (1)
⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) + \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 0 ………….. (i)
Now, u = x^y log u = log x^y = y log x
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\)log u = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(y log x)
⇒ \(\frac{1}{\mathrm{u}}\frac{\mathrm{du}}{\mathrm{dx}}\) = y\(\frac{\mathrm{d}}{\mathrm{dx}}\)log x + log x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y\(\frac{1}{\mathrm{x}}\) + log x \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
∴ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = \(u\left(\frac{y}{x}+\log x \cdot \frac{d y}{d x}\right)\)

Question 12.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of the function y^x = x^y.
Solution:
Given that y^x = x^y ⇒ x^y = y^x
Taking logarithms,
log x^y = log y^x
⇒ y log x = x log y.
Differentiating w.r.t. x we have

Question 13.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of the function (cos x)^y = (cos y)^x
Solution:
Given that (cos x)^y = (cos y)^x ⇒ log(cos x)^y = log(cos y)^x
⇒ y log cos x = x log cos y.
Differentiating both sides w.r.t. x, we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (y log cos x) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x log cos y)
Applying Product Rule on both sides,
y\(\frac{\mathrm{d}}{\mathrm{dx}}\)log cos x + log cos x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x\(\frac{\mathrm{d}}{\mathrm{dx}}\)log cos y + log ços y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x
⇒ y. \(\frac{1}{\cos x}\frac{\mathrm{d}}{\mathrm{dx}}\) cos x + log cos x\(\frac{1}{\cos y}\frac{\mathrm{dy}}{\mathrm{dx}}\) = x. \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos y + log cos y
⇒ y \(\frac{1}{\cos x}\) (-sin x) + log cos x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x \(\frac{1}{\cos y}\) (-sin y \(\frac{\mathrm{dy}}{\mathrm{dx}}\)) + log cos y
⇒ -y tan x + log cos x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -x tan y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + log cos y
⇒ x tan y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + log cos x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y tan x + log cos y
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\)(x tan y + log cos x) = y tan x + log cos y
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}\)

Question 14.
Find the derivative of the function given by f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f'(1).
Solution:
Given f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) …………. (i)
Taking logs on both sides, we have
log f(x) = log(1 + x) + log(1 + x²) + log (1 + x⁴) + log(1 + x⁸)
Differentiating both sides w.r.t. x, we have

Question 15.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three was mentioned below:
i) by using product rule
ii) b expanding the product to obtain a single polynomial.
iii) by logarithmic differentiation. l)o they all give the same answer?
Solution:
Let y = (x² – 5x + 8) (x³ + 7x + 9) ……………… (1)
(i) To find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) by using Product Rule:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (x² – 5x + 8)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x³ + x + 9) + (x³ + 7x + 9)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x² – 5x + 8)
= (x² – 5x + 8x)(3x² + 7) + (x³ + 7x + 9)(2x – 5)
= 3x⁴ + 7x² – 15x³ – 35x + 24x² + 56 + 2x⁴ – 5x³ + 14x² – 35x + 18x – 45
= 5x⁴ – 20x³ + 45x² – 52x + 11 …………… (2)

(ii) To find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) by expanding the product to obtain a single polynomial:
From (1)y = (x² – 5x + 8)(x³ + 7x + 9)
= x⁵ + 7x³ + 9x² – 5x⁴ – 35x² – 45x + 8x³ + 56x + 72
⇒ y = x⁵ – 5x⁴ + 15x³ – 26x² + 11x + 72
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 5x⁴ – 20x³ + 45x² – 52x + 11 ……………….. (3)

(iii) To find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) by logarithmic differentiation:
Taking logs on both sides of (1), we have
log y = log(x² – 5x + 8)(x³ + 7x + 9)
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\)log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)log(x² – 5x + 8) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) log(x³ + 7x + 9)

= 5x⁴ – 20x³ + 45x² – 52x + 11
From (2), (3) and (4) we get the same answer in all the 3 ways

Question 16.
If u, y and w are functions of x, then show that
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (uvw) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) v.w + u.\(\frac{\mathrm{dv}}{\mathrm{dx}}\) .w + u.v\(\frac{\mathrm{dw}}{\mathrm{dx}}\) in two ways – first by repeated application of product rule, second b logarithm differentiation.
Solution:
Given that u, y and w are functions of x.
To prove: \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uvw) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) v.w + u.\(\frac{\mathrm{dv}}{\mathrm{dx}}\) .w + u.v\(\frac{\mathrm{dw}}{\mathrm{dx}}\) …………. (i)
(i) To prove eqn. (i): by using product rule
L.H.S = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(uvw) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(uv)w = uv\(\frac{\mathrm{d}}{\mathrm{dx}}\)(w) + w\(\frac{\mathrm{d}}{\mathrm{dx}}\)(uv)
Again Applying Product Rule on \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uv)
L.H.S. = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(uvw) = uvv\(\frac{\mathrm{dw}}{\mathrm{dx}}\) + w[u\(\frac{\mathrm{d}}{\mathrm{dx}}\)v + v\(\frac{\mathrm{d}}{\mathrm{dx}}\)u] = uv\(\frac{\mathrm{dw}}{\mathrm{dx}}\) + uw\(\frac{\mathrm{dv}}{\mathrm{dx}}\) + vw\(\frac{\mathrm{du}}{\mathrm{dx}}\) = R.H.S
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uvw) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) vw + u\(\frac{\mathrm{dv}}{\mathrm{dx}}\)w + uv\(\frac{\mathrm{dw}}{\mathrm{dx}}\). Hence proved.

(ii) Let y = uvw
Taking logs on both sides logy = log( uvw) = log u + log v + log w
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\)log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)log u + \(\frac{\mathrm{d}}{\mathrm{dx}}\) log v + \(\frac{\mathrm{d}}{\mathrm{dx}}\) log w
⇒ \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x} \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right]\)
Putting y = uvw, \(\frac{\mathrm{d}}{\mathrm{dx}}\)(uvw) = uvw \(\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right)\)
= \(\frac{\mathrm{du}}{\mathrm{dx}}\) vw + u\(\frac{\mathrm{dv}}{\mathrm{dx}}\)w + uv\(\frac{\mathrm{dw}}{\mathrm{dx}}\)
Hence proved.
∴ In both the methods, we will get the same answer

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7h Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7h

I.

Question 1.
Evaluate \(\int_{-1}^1(x+1) d x\)
Solution:
Let I = \(\int_{-1}^1(x+1) d x=\left[\frac{x^2}{2}+x\right]_{-1}^1=\left[\frac{1}{2}+1\right]-\left[\frac{1}{2}-1\right]=\frac{1}{2}+1-\frac{1}{2}+1=2\)

Question 2.
Evaluate \(\int_2^3 \frac{1}{x} d x\)
Solution:
Let I = \(\int_2^3 \frac{1}{x} d x=[\log |x|]_2^3=\log |3|-\log |2|=\log \frac{3}{2}\)

Question 3.
Evaluate \(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
Solution:
\(\int_1^2\left[4 x^3-5 x^2+6 x+9\right] d x=\left[4 \frac{x^4}{4}-5 \frac{x^3}{3}+6 \frac{x^2}{2}+9 x\right]_1^2\)
= \(\left[x^4-\frac{5}{3} x^3+3 x^2+9 x\right]_1^2=\left[2^4-\frac{5}{3}(2)^3+3(2)^2+9(2)\right]-\left[1-\frac{5}{3}+3+9\right]\)
= \(\left[16-\frac{40}{3}+12+18\right]-\left[13-\frac{5}{3}\right]=\left[46-\frac{40}{3}\right]-\left[13-\frac{5}{3}\right]\)
= \(46-\frac{40}{3}-13+\frac{5}{3}=33-\frac{40}{3}+\frac{5}{3}=\frac{99-40+5}{3}=\frac{104-40}{3}=\frac{64}{3}\).

Question 4.
Evaluate \(\int_0^\pi 4 \sin 2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{4}} \sin 2 x d x=\left[\frac{-\cos 2 x}{2}\right]_0^{\frac{\pi}{4}}=\left[\frac{-\cos 2 \frac{\pi}{4}}{2}\right]-\left[\frac{-\cos 0}{2}\right]=0-\left(\frac{-1}{2}\right)=0+\frac{1}{2}=\frac{1}{2}\) [∵ \(\cos \frac{\pi}{2}\) = 0]

Question 5.
Evaluate \(\int_0^\pi 2 \cos 2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{2}} \cos 2 x d x=\left[\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}=\frac{\sin \pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}=0\) [∵ sin π = sin 180° = 0]

Question 6.
Evaluate \(\int_4^5 e^x d x\)
Solution:
Let I = \(\int_4^5 e^x d x=\left[e^x\right]_4^5\) = e⁵ – e⁴ = e⁴(e – 1)

Question 7.
Evaluate \(\int_0^\pi 4 \tan x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \tan x d x=[-\log \cos x]_0^{\pi / 4}=-\log \left|\cos \frac{\pi}{4}\right|+\log |\cos 0|\)
= \(-\log \left|\frac{1}{\sqrt{2}}\right|+\log |1|=-\log (2)^{\frac{-1}{2}}=\frac{1}{2} \log 2\)

Question 8.
Evaluate \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x\)
Solution:
I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x=[\log |{cosec} x-\cot x|]_{\pi / 6}^{\pi / 4}=\log \left|{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|{cosec} \frac{\pi}{6}-\cot \frac{\pi}{6}\right|\)
= \(\log |\sqrt{2}-1|-\log |2-\sqrt{3}|=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Question 9.
Evaluate \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1=\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{\pi}{2}-0=\frac{\pi}{2}\)

Question 10.
Evaluate \(\int_0^1 \frac{d x}{1+x^2}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1=\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}\)

Question 11.
Evaluate \(\int_2^3 \frac{d x}{x^2-1}\)
Solution:
Let I = \(\int_2^3 \frac{\mathrm{dx}}{\mathrm{x}^2-1}=\left[\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|\right]_2^3\)
= \(\frac{1}{2}\left[\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|\right]=\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]=\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{3}\right]=\frac{1}{2}\left[\log \frac{3}{2}\right]\)

Question 12.
Evaluate \(\int_2^3 \frac{x d x}{x^2+1}\)
Solution:
Let I = \(\int_2^3 \frac{x}{x^2+1} d x=\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x=\frac{1}{2}\left[\log \left(1+x^2\right)\right]_2^3\)
= \(\frac{1}{2}\left[\log \left(1+3^2\right)-\log \left(1+2^2\right)\right]=\frac{1}{2}[\log (10)-\log (5)]=\frac{1}{2} \log \left(\frac{10}{5}\right)=\frac{1}{2} \log 2\)

Question 13.
Evaluate \(\int_0^1 x e^{x^2} d x\)
Solution:
Let I = \(\int_0^1 x e^{x^2} d x\) put x² = t ⇒ 2x dx = dt
As x → 0, t → 0 and as x → 1, t → 1
∴ I = \(\frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1=\frac{1}{2} e-\frac{1}{2} e^0=\frac{1}{2}(e-1)\)

Question 14.
Evaluate \(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x\)
Solution:
\(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x=\left[2 \tan x+\frac{x^4}{4}+2 x\right]_0^{\pi / 4}\)
= \(\left[\left(2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^4+2\left(\frac{\pi}{4}\right)\right)-(2 \tan 0+0+0)\right]=2+\frac{\pi^4}{4^5}+\frac{\pi}{2}=2+\frac{\pi}{2}+\frac{\pi^4}{1024}\)

Question 15.
Evaluate \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x\)
Solution:
Let I = \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x=-\int_0^\pi\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right) d x\)
= \(-\int_0^\pi \cos x d x=-[\sin x]_0^\pi\) = -(sin π – sin 0) = 0 – 0 = 0

Question 16.
Evaluate \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x=\int_0^{\pi / 2}\left(\frac{1+\cos 2 x}{2}\right) d x=\left[\frac{x}{2}+\frac{\sin 2 x}{4}\right]_0^{\pi / 2}=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^{\pi / 2}\)
= \(\frac{1}{2}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(0+\frac{\sin 0}{2}\right)\right]=\frac{1}{2}\left[\frac{\pi}{2}+0-0-0\right]=\frac{\pi}{4}\)

Question 17.
Find the integral of \(\int_0^1 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{x}}}\)
Solution:

Question 18.
Evaluate \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
I = \(\int_0^{\frac{\pi}{2}} \sin ^2 x \cdot \sin x d x=\int_0^{\frac{\pi}{2}}\left(1-\cos ^2 x\right) \sin x d x=\int_0^{\frac{\pi}{2}} \sin x d x-\int_0^{\frac{\pi}{2}} \cos ^2 x \cdot \sin x d x\)
= \([-\cos x]_0^{\frac{\pi}{2}}+\left[\frac{\cos ^3 x}{3}\right]_0^{\frac{\pi}{2}}=\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[\cos ^3 \frac{\pi}{2}-\cos ^3 0\right]=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}\)

II.

Question 1.
Evaluate \(\int_0^1 \frac{2 x+3}{5 x^2+1} d x\)
Solution:

Question 2.
Evaluate \(\int_0^2 \frac{6 x+3}{x^2+4} d x\)
Solution:

Question 3.
Evaluate \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\)
Solution:
Let I = \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
Dividing the Nr. & Dr. By x² + 4x + 3 and simplifying, we get
Dividing 5x² by x² + 4x + 3, we get
I = \(\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right] d x=\int_1^2 5 d x-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x=[5 x]_1^2-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x\) …(1)
Let I₁ = \(\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x\) …………(1)
Let 20x + 15 = A\(\frac{d}{d x}\)(x² + 4x + 3) + B = 2Ax + (4A + B)
Equating the coefficients of x and constant term, we get
A = 10 and B = -25
∴ 20x + 15 = 10(2x + 4) – 25 [∵ \(\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\)]
⇒ I₁ = \(10 \int_1^2 \frac{2 x+4}{x^2+4 x+3} d x-25 \int_1^2 \frac{d x}{(x+2)^2-1^2}=10\left[\log \left|x^2+4 x+3\right|\right]_1^2-25\left[\frac{1}{2} \log \left(\frac{x+2-1}{x+2+1}\right)\right]_1^2\)
= \(\left[10 \log \left(x^2+4 x+3\right)\right]_1^2-25\left[\frac{1}{2} \log \left(\frac{x+1}{x+3}\right)\right]_1^2\)
= [10log15 – 10log8] – 25 \(\left[\frac{1}{2} \log \frac{3}{5}-\frac{1}{2} \log \frac{2}{4}\right]\)
= [10 log(5 × 3) – 10 log(4 × 2)] – \(\frac{25}{2}\)[log 3 – log 5 – log 2 + log 4]
= [10 log5 + 10log3 – 10log4 – 10log2] – \(\frac{25}{2}\)[log3 – log5 – log2 + log4]
= \(\left[10+\frac{25}{2}\right] \log 5+\left[-10-\frac{25}{2}\right] \log 4+\left[10-\frac{25}{2}\right] \log 3+\left[-10+\frac{25}{2}\right] \log 2\)
= \(\frac{45}{2} \log 5-\frac{45}{2} \log 4-\frac{5}{2} \log 3+\frac{5}{2} \log 2=\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\)
Substituting the value I₁, in (1), we get
I = \(5-\left[\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\right]=5-\frac{5}{2}\left[9 \log \frac{5}{4}-\log \frac{3}{2}\right]\)

Question 4.
Evaluate \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right] d x\)
Solution:
Let I = \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right] d x\)

Question 5.
Evaluate \(\int_1^3 \frac{d x}{x^2(x+1)}\)
Solution:
Let, \(\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \Rightarrow \frac{A x(x+1)+B(x+1)+C\left(x^2\right)}{x\left(x^2\right)(x+1)}\)
⇒ 1 = Ax(x + 1) + B(x + 1) + C(x²) ⇒ 1 = Ax² + Ax + Bx + B + Cx²
Equatingthe coefficients of x², x and constant terms, we get A + C = 0, A + B = 0, B = 1
On solving these equations, we get A = -1, C = 1, B = 1
∴ \(\frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\)
⇒ I = \(\int_1^3\left[-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\right] d x=\left[-\log x-\frac{1}{x}+\log (x+1)\right]_1^3\)
= \(\left[\log \left(\frac{x+1}{x}\right)-\frac{1}{x}\right]_1^3=\log \left(\frac{4}{3}\right)-\frac{1}{3}-\log \left(\frac{2}{1}\right)+1\)
= log 4 – log 3 – log 2 + \(\frac{2}{3}\) = log 2 – log 3 + \(\frac{2}{3}\) = \(\log \left(\frac{2}{3}\right)+\frac{2}{3}\), Hence proved.

Question 6.
Evaluate \(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}} d x\)
Solution:
Put x^1/4 = t ⇒ x = t⁴ ⇒ dx = 4t³ dt.
Also, when x = 0 we get t = 0 and when x = 1 we get t = 1^1/4 = 1

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5d

Question 1.
Differentiate \(\frac{e^x}{\sin x}\) w.r.t. x
Solution:

Question 2.
Differentiate e^sin-1x w.r.t. x
Solution:
Let y = e^sin-1x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^sin-1x \(\frac{\mathrm{d}}{\mathrm{dx}}\)sin^-1x [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= e^sin-1x \(\frac{1}{\sqrt{1-x^2}}\), x ∈ (-1, 1)

Question 3.
Differentiate e^x3 w.r.t. x.
Solution:
Let y = e^x3 = e^(x3)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^(x3) \(\frac{\mathrm{d}}{\mathrm{dx}}\)x³ [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= e^(x3)3x² = 3x²e^(x3)

Question 4.
Differentiate sin(tan^-1e^-x) w.r.t. x:
Solution:
Let y = sin(tan^-1e^-x)

Question 5.
Differentiate log(cos e^x) w.r.t. x.
Solution:
Let y = log(cos e^x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{\cos e^x} \frac{d}{d x}\) (cos e^x)
= \(\frac{1}{\cos \mathrm{e}^{\mathrm{x}}}\)(-sin e^x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^x
= -(tan e^x)e^x = -e^x (tan e^x)

Question 6.
Differentiate e^x + e^x2+ ……….. + e^x5 w.r.t. x.
Solution:
Let y = e^x + e^x2+ e^x3 + e^x4 + e^x5
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x2+ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^x3 + \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^x4 + ……….. + \(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x5
= e^x + e^x2\(\frac{\mathrm{d}}{\mathrm{dx}}\) x² + e^x3 \(\frac{\mathrm{d}}{\mathrm{dx}}\) x³ + e^x4\(\frac{\mathrm{d}}{\mathrm{dx}}\) x⁴ + ……….. + e^x5 \(\frac{\mathrm{d}}{\mathrm{dx}}\) x⁵ [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= e^x + e^x2 2x + e^x3 3x² + e^x4 4x³ + e^x5 5x⁴
= e^x + 2x e^x2+ 3x² e^x3 + 4x³ e^x4 + 5x⁴ e^x5

Question 7.
Differentiate \(\sqrt{e^{\sqrt{x}}}\), x > 0 w.r.t. x.
Solution:
Let y = \(\sqrt{e^{\sqrt{x}}}\) = \(\left(e^{\sqrt{x}}\right)^{1 / 2}\)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{2}\left(e^{\sqrt{x}}\right)^{-1 / 2} \frac{d}{d x} e^{\sqrt{x}}\)
= \(\frac{1}{2 \sqrt{e^{\sqrt{x}}}} e^{\sqrt{x}} \frac{d}{d x} \sqrt{x}\) = \(\frac{1}{2 \sqrt{e^{\sqrt{x}}}} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}\)
= \(\frac{e^{\sqrt{x}}}{4 \sqrt{x} \sqrt{e^{\sqrt{x}}}}\)=\(\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}\)

Question 8.
Differentiate log (log x), x > 1 w.r.t. x
Solution:
Let y = log (log x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{\log x} \frac{d}{d x}(\log x)\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) logf(x) = \(\frac{1}{f(x)} \frac{d}{d x}\) f(x)]
= \(\frac{1}{\log x} \frac{d}{d x}(\log x)\)
= \(\frac{1}{x \log x}\)

Question 9.
Differentiate \(\frac{\cos x}{\log x}\), x > 0 w.r.t. x
Solution:

Question 10.
Differentiate cos(logx + e^x), x > 0 w.r.t x.
Solution:
Let y = cos(logx + e^x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin(log x + e^x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(log x + e^x)
= -sin(log x + e^x) . (\(\frac{1}{x}\) + e^x) = -(\(\frac{1}{x}\) + e^x)sin(log x + e^x)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7g Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7g

I.

Question 1.
Find the integral of \(\sqrt{4-x^2}\)
Solution:
Let I = \(\int \sqrt{4-x^2} d x=\int \sqrt{(2)^2-(x)^2} d x\) [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]
∴ I = \(\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}+C=\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1} \frac{x}{2}+C\)

Question 2.
Find the integral of \(\sqrt{1-4 x^2}\)
Solution:
Let I = \(\int \sqrt{1-4 \mathrm{x}^2} \mathrm{dx}=\int \sqrt{1^2-(2 \mathrm{x})^2} \mathrm{dx}\) dx Put 2x = t ⇒ 2 dx = dt
∴ I = \(\frac{1}{2} \int \sqrt{1^2-t^2}=\frac{1}{2}\left[\frac{t}{2} \sqrt{1-t^2}+\frac{1}{2} \sin ^{-1} t\right]+C=\frac{t}{4} \sqrt{1-t^2}+\frac{1}{4} \sin ^{-1} t+C\)
= \(\frac{2 x}{4} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+C=\frac{x}{2} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+C\)

Question 3.
Find the integral of \(\sqrt{x^2+4 x+6}\)
Solution:

Question 4.
Find the integral of \(\sqrt{x^2+4 x+1}\)
Solution:
Let I = \(\int \sqrt{x^2+4 x+1} d x=\int \sqrt{\left(x^2+4 x+4\right)-3} d x\)
= \(\int \sqrt{(x+2)^2-(\sqrt{3})^2} d x\) [∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C\)]
∴ I = \(\frac{(x+2)}{2} \sqrt{x^2+4 x+1}-\frac{3}{2} \log \left|(x-2)+\sqrt{x^2+4 x+1}\right|+C\)

Question 5.
Find the integral of \(\sqrt{1-4 x-x^2}\)
Solution:
Let I = \(\int \sqrt{1-4 x-x^2} d x=\int \sqrt{1-\left(x^2+4 x+4-4\right)} d x=\int \sqrt{1+4-(x+2)^2} d x\)
= \(\int \sqrt{(\sqrt{5})^2-(x+2)^2} d x\) [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]
∴ I = \(\frac{(x+2)}{2} \sqrt{1-4 x-x^2}+\frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C\)

Question 6.
Find the integral of \(\sqrt{x^2+4 x-5}\)
Solution:
Let I = \(\int \sqrt{x^2+4 x-5} d x\)
= \(\int \sqrt{\left(x^2+4 x+4\right)-9} d x=\int \sqrt{(x+2)^2-3^2} d x\)
∴ I = \(\frac{(x+2)}{2} \sqrt{x^2+4 x-5}-\frac{9}{2} \log \left|(x+2)+\sqrt{x^2+4 x-5}\right|+C\)

Question 7.
Find the integral of \(\sqrt{1+3 x-x^2}\)
Solution:

Question 8.
Find the integral of \(\sqrt{x^2+3 x}\)
Solution:

Question 9.
Find the integral of \(\sqrt{1+\frac{x^2}{9}}\)
Solution: