Srinivas

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:

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

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

I.

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of 2x + 3y = sin x
Solution:
Given that 2x + 3y = sin x.
Differentiating both sides w.r.t x, we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (3y) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x
∴ 2 + 3 \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\cos x-2}{3}\)
⇒ 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x – 2

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of 2x + 3y = sin y
Solution:
Given that 2x + 3y = sin y ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (3y)
= \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin y
∴ 2 + 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) (cos y – ) = 2
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(=\frac{2}{\cos y-3}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of ax + by² = cos y
Solution:
Given that ax + by² = cos y ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (ax) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (by²) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos y)
∴ a + b.2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin y\(\frac{\mathrm{d}}{\mathrm{dx}}\)
⇒ 2by \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + sin y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -a
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) (2by + sin y) = -a ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-a}{2 b y+\sin y}\)

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of xy + y² = tan x + y
Solution:
Given that xy + y² = tan x + y. ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(xy) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y² = \(\frac{\mathrm{d}}{\mathrm{dx}}\)tan x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y
(x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y + y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec²x + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) [∵ From Product Rule]
⇒ x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec² x + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) ⇒ x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec² x – y
⇒ (x + 2y – 1)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec²x – y
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sec ^2 x-y}{x+2 y-1}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x² + xy + y² = 100
Solution:
Given that ² + xy + y² = 100
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)x² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)xy + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y² = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(100)
∴ 2x + (x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y + y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ 2x + x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ (x + 2y)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2x – y
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(2 x+y)}{x+2 y}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x³ + x²y + xy² + y³ = 81
Solution:
Given that x³ + x²y + xy² + y³ = 81
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)x³ + \(\frac{\mathrm{d}}{\mathrm{dx}}\)x²y + \(\frac{\mathrm{d}}{\mathrm{dx}}\)xy² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y³ = 81
∴ 3x² + x² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y.\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²) + x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y² + y² \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x) = 3y²\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ 3x² + x² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y2x + x2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y² . 1 + 3y² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² +2xy + 3y²) = -3x² – 2xy – y²
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\left(3 x^2+2 x y+y^2\right)}{x^2+2 x y+3 y^2}\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of sin²y + cos xy = k
Solution:
Given that sin² y + cos xy = k
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin y)² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)cos xy = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(k)
∴ 2(sin.y)\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin y – sin xy\(\frac{\mathrm{d}}{\mathrm{dx}}\)(xy) = 0
⇒ 2sin y cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – sin xy(x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y.1) = 0
⇒ sin2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – x sin xy\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y sin xy = 0
⇒ (sin2y – x sin xy)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y sin xy
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y \sin x y}{\sin 2 y-x \sin x y}\)

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of sin²x + cos² y = 1
Solution:
Given that sin² x + cos² y = 1
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x)² + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos y)² = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (1)
∴ 2(sin x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x + 2(cos y)\(\frac{\mathrm{d}}{\mathrm{dx}}\) cos y = 0
⇒ 2 sin x cos x + 2cos y(-sin y\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = 0
⇒ 2 sin x cos x – 2 sin y cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ sin 2x – sin 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = o
⇒ sin 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin 2x
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin 2 x}{\sin 2 y}\)

II.

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Solution:
Given that y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
To simplify the given inverse t-function, we put x = tanθ.
∴ y = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(sin 2θ) = 2θ
⇒ y = 2tan^-1x (∵ x tanθ ⇒ θ = tan^-1x)
∴ \(\frac{d y}{d x}=2 \frac{1}{1+x^2}=\frac{2}{1+x^2}\)

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\), \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Solution:
Given y = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\), \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Put x = tanθ
∴ y = tan^-1\(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\) = tan^-1(tan 3θ) = 3θ
⇒ y = 3 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3 . \(\frac{1}{1+x^2}=\frac{3}{1+x^2}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Solution:
Given that y = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Put x = tanθ
∴ y = cos^-1\(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) = cos^-1(cos 2θ)
y = 2θ = 2tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2 . \(\frac{1}{1+x^2}=\frac{2}{1+x^2}\)

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sin^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Solution:
Given that y = sin^-1\(\left(\frac{1-x^2}{1+x^2}\right)\)
Put x = tanθ
∴ y = sin^-1\(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(cos 2θ)
= sin^-1 sin(\(\frac{\pi}{2}\) – 2θ) = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 – 2\(\frac{1}{1+x^2}=\frac{-2}{1+x^2}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = cos^-1\(\left(\frac{2 x}{1+x^2}\right)\), -1 < x < 1
Solution:
Given that y = cos^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Put x = tanθ
∴ y = cos^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = cos^-1(sin 2θ)
= cos^-1 sin(\(\frac{\pi}{2}\) – 2θ) = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 – 2\(\frac{1}{1+x^2}=\frac{-2}{1+x^2}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sin^-1(2x\(\sqrt{1-\mathrm{x}^2}\)), \(-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
Solution:
Given that y = sin^-1(2x\(\sqrt{1-\mathrm{x}^2}\)). Put x = sinθ
∴ y = sin^-1(2 sinθ\(1-\sin ^2 \theta\)) = sin^-1(2sinθ\(\cos ^2 \theta\)) = sin^-1(2 sinθ cosθ)
⇒ y = sin^-1 (sin 2θ) = 2θ = 2 sin^-1 x [∵ x = sin θ ⇒ θ = sin^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sec^-1\(\left(\frac{1}{2 x^2-1}\right),\), 0 < x < \(\frac{1}{\sqrt{2}}\)
Solution:
Put x = cosθ then 2x² – 1 = 2cos²θ – 1 = cos 2θ) = 2θ
∴ y = sec^-1\(\left(\frac{1}{\cos 2 \theta}\right)\) = sec^-1 (sec 2θ) = 2θ = 2 cos^-1
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2cos^-1x)
= 2\(\left(\frac{-1}{\sqrt{1-x^2}}\right)=\frac{-2}{\sqrt{1-x^2}}\)

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

AP Inter 2nd Year Maths Integrals Solutions Exercise 7f

I.

Question 1.
Find the integral of x sin x
Solution:
Let I = ∫x sin x dx
Taking u = x, v = sinx and integrating by parts, we have
I = \(x \int \sin x d x-\int\left[\left(\frac{d}{d x}(x)\right) \int \sin x d x\right] d x=x(-\cos x)-\int 1 \cdot(-\cos x) d x\) = -x cosx + sinx + C

Question 2.
Find the integral of x sin 3x
Solution:
Let I = ∫x sin 3x dx
Taking u = x, v = sin3x and integrating by parts, we have
I = \(x \int \sin 3 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sin 3 x d x\right] d x\)
= \(x\left(\frac{-\cos 3 x}{3}\right)-\int 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x=\frac{-x \cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x\)
= \(\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x+C=\frac{-x}{3} \cos 3 x+\frac{1}{9} \sin 3 x+C\)

Question 3.
Find the integral of x log x
Solution:
Let I = ∫x log x dx
Taking u = log x, v = sin3x and integrating by parts, we have
I = \(\log x \int x d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x d x\right] d x\)
= \(\log x \cdot \frac{x^2}{2}-\int \frac{1}{x} \cdot \frac{x^2}{2} d x=\frac{x^2 \log x}{2}-\int \frac{x}{2} d x=\frac{x^2 \log x}{2}-\frac{x^2}{4}+C\)

Question 4.
Find the integral of x log 2x
Solution:
Let I = ∫x log 2x dx
Taking u = log 2x, v = sin3x and integrating by parts, we have
I = \(\log 2 x \int x d x-\int\left[\left(\frac{d}{d x} \log 2 x\right) \int x d x\right] d x\)
= \(\log 2 x \cdot \frac{x^2}{2}-\int \frac{2}{2 x} \cdot \frac{x^2}{2} d x=\frac{x^2 \log 2 x}{2}-\int \frac{x}{2} d x=\frac{x^2 \log 2 x}{2}-\frac{x^2}{4}+C\)

Question 5.
Find the integral of x² log x
Solution:
Let I = ∫x² log x dx
Taking u = log x, v = x² and integrating by parts, we have
I = \(\log x \int x^2 d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x^2 d x\right] d x\)
= \(\log x .\left(\frac{x^3}{3}\right)-\int \frac{1}{x} . \frac{x^3}{3} d x=\frac{x^3 \log x}{3}-\int \frac{x^2}{3} d x=\frac{x^3 \log x}{3}-\frac{x^3}{9}+C\)

Question 6.
Find the integral of x sec² x
Solution:
Let I = ∫x sec² x dx
Taking u = x, v = sec²x and integrating by parts, we have
I = \(x \int \sec ^2 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sec ^2 x d x\right] d x\)
= x tan x – ∫1. tan xdx = x tanx + log|cos x| + C

Question 7.
Find the integral of tan^-1 x
Solution:
Let I = ∫1.tan^-1 x dx
Taking u = tan^-1 x, v = 1 and integrating by parts, we have
I = \(\tan ^{-1} x \int 1 d x-\int\left[\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 . d x\right] d x\)
= \(\tan ^{-1} x . x-\int \frac{1}{1+x^2} x d x=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^2} d x\)
= \(x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^2\right|+C=x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^2\right)+C\)

Question 8.
Find the integral of (x² + 1)log x
Solution:
Let I = ∫(x² + 1)log x dx = ∫x² logxdx + ∫logx dx
Let I = I₁ + I₂ ……….(1)
Where, I₁ = ∫x² logxdx and I₁ = ∫logx dx
I₁ = ∫x²log xdx
Taking u = log x, v = x² and integrating by parts, we have
I₁ = \(\log x \int x^2 d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x^2 d x\right] d x=\log x \cdot \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x\)
= \(\frac{x^3}{3} \log x-\frac{1}{3}\left(\int x^2 d x\right)=\frac{x^3}{3} \log x-\frac{x^3}{9}+C_1\) …(2)
I₂ = ∫log xdx
Taking u = log x, v = 1 and integrating by parts, we have
I₂ = \(\log x \int 1 . d x-\int\left[\left(\frac{d}{d x} \log x\right) \int 1 . d x\right]\)
= \(\log x \cdot x-\int \frac{1}{x} \cdot x d x=x \log x-\int 1 \cdot d x=x \log x-x+C_2\) …………(3)
Using equation (2) and (3) in (1), we get

Question 9.
Find the integral of e^x(sinx + cosx)
Solution:
We know that ∫e^x{f(x) + f(x)} dx = e^xf(x) + C
Here f(x) = sin x and f'(x) = cos x
∴ I = ∫e^x(sin x + cos x) dx = e^xsin x + C

Question 10.
Find the integral of e^x\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)
Solution:
Let I = ∫e^x\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)dx
We know that ∫e^x[f(x) + f'(x)]dx = e^xf(x) + C
Here, f(x) = \(\frac{1}{x}\) & f'(x) = \(\frac{-1}{x^2}\)
∴ I = \(e^x\left(\frac{1}{x}\right)+C=\frac{e^x}{x}+C\)

II.

Question 1.
Find the integral of x²e^x
Solution:
Let I = ∫ x²e^x dx
Taking u = x², v = e^x and integrating by parts, we have
I = \(x^2 \int e^x d x-\int\left[\left(\frac{d}{d x} x^2\right) \int e^x d x\right] d x=x^2 e^x-\int 2 x e^x d x=x^2 e^x-2 \int x e^x d x\)
Again using integration by parts, we have
I = \(x^2 e^x-2\left[x \int e^x d x-\int\left(\frac{d}{d x} x\right) \int e^x d x\right] d x\)
= x²e^x – 2[xe^x – e^x dx] = x²e^x – 2[xe^x – ∫e^x]
= x²e^x – 2xe^x + 2e^x + C = e^x(x² – 2x + 2) + C

Question 2.
Find the integral of x sin^-1 x
Solution:
Let I = ∫ xsin^-1x dx
Taking u = sin^-1 x, v = x and integrating by parts we have
I = \(\sin ^{-1} x \int x d x-\int\left[\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right] d x=\sin ^{-1} x\left(\frac{x^2}{2}\right)-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} d x\)
= \(\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} d x\)

Question 3.
Find the integral of x tan^-1 x
Solution:
Let I = ∫ x tan^-1 xdx
Taking u = tan^-1 x, v = x and integrating by parts we have

Question 4.
Find the integral of x cos^-1 x
Solution:
Let I = ∫ x cos^-1 xdx
Taking u = cos^-1 x, v = x and integrating by parts we have

Question 5.
Find the integral of \(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} \cdot \cos ^{-1} x d x\)
Taking u = cos^-1 x, v = \(\left(\frac{-2 x}{\sqrt{1-x^2}}\right)\) and integrating by parts we have

Question 6.
Find the integral of x(log x)²
Solution:
Let I = ∫ x(log x)² dx
Taking u = (log x)², v = x and integrating by parts, we have
I = \((\log x)^2 \int x d x-\int\left[\left(\frac{d}{d x}(\log x)^2\right) \int x d x\right] d x\)
= \(\frac{x^2}{2}(\log x)^2-\left[\int 2 \log x \frac{1}{x} \frac{x^2}{2} d x\right]=\frac{x^2}{2}(\log x)^2-\int x \log x d x\)
Again, using integrated by parts, we have
I = \(\frac{x^2}{2}(\log x)^2-\left[\log x \int x d x-\int\left(\left(\frac{d}{d x} \log x\right) \int x d x\right) d x\right]\)
= \(\frac{x^2}{2}(\log x)^2-\left[\frac{x^2}{2} \log x-\int \frac{1}{x} \cdot \frac{x^2}{2} d x\right]=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2} \log x+\frac{1}{2} \int x d x\)
= \(\frac{x^2}{2}(\log x)^2-\frac{x^2}{2} \log x+\frac{x^2}{4}+C\)

Question 7.
Find the integral of \(\frac{x e^x}{(1+x)^2}\)
Solution:
We know that ∫e^x[f(x) + f'(x)] dx = e^xf(x) + C
Here, f(x) = \(\frac{1}{1+x}\) and f'(x) = \(\frac{-1}{(1+x)^2}\)
∴ I = \(\int \frac{x e^x}{(1+x)^2} d x=\int e^x\left[\frac{x}{(1+x)^2}\right] d x=\int e^x\left[\frac{1+x-1}{(1+x)^2}\right] d x=\int e^x\left[\frac{1}{1+x}-\frac{1}{(1+x)^2}\right] d x\)
∴ I = \(e^x \frac{1}{1+x}+C=\frac{e^x}{1+x}+C\)

Question 8.
Find the integral of \(e^x\left(\frac{1+\sin x}{1+\cos x}\right)\)
Solution:

[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x) + C Here \(\tan \frac{x}{2}\) = f(x) & f'(x) = \(\frac{1}{2}\)sec²\(\frac{x}{2}\)]

Question 9.
Find the integral of \(\frac{(x-3) e^x}{(x-1)^3}\)
Solution:
Let I = \(\int e^x\left[\frac{x-3}{(x-1)^3}\right] d x=\int e^x\left[\frac{x-1-2}{(x-1)^3}\right] d x=\int e^x\left[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\right] d x\)
Here f(x) = \(\frac{1}{(x-1)^2}\) and f'(x) = \(\frac{-2}{(x-1)^3}\)
We know that ∫ e^x[f(x) + f'(x)] dx = e^xf(x) + C ∵ I = \(\frac{\mathrm{e}^{\mathrm{x}}}{(\mathrm{x}-1)^2}\) + C

Question 10.
Find the integral of e^2x sin x
Solution:
Let I = e^2x sin x dx …….(1)
Taking u = sin x, v = e^2x and integrating by parts, we have

Question 11.
Find the integral of \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Solution:
Let x = tan θ ⇒ dx = sec² θdθ
∴ sin^-1\(\left(\frac{2 x}{1+x^2}\right)\) = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(sin 2θ) = 2θ
∫ sin^-1\(\left(\frac{2 x}{1+x^2}\right)\) dx = ∫ 2θ.sec² θdθ = 2∫θ. sec² θdθ
Using integration by parts, we get
I = \(\left[\theta \cdot \int \sec ^2 \theta \mathrm{~d} \theta-\int\left[\left(\left(\frac{\mathrm{d}}{\mathrm{~d} \theta} \theta\right) \int \sec ^2 \theta \mathrm{~d} \theta\right)\right] \mathrm{d} \theta=2\left[\theta \cdot \tan \theta-\int \tan \theta \mathrm{d} \theta\right]\right.\)
= 2[θ. tanθ + log |cos θ|] + C = \(2\left[x \tan ^{-1} x+\log \left|\frac{1}{\sqrt{1+x^2}}\right|\right]+C\)
= 2x tan^-1 x + 2log(1 + x²)^{\(\frac{-1}{2}\)} + C = \(2 x \tan ^{-1} x+2\left[\frac{-1}{2} \log \left(1+x^2\right)\right]+C\)
= 2x tan^-1 x – log(1 + x²) + C

Question 12.
Find the integral of \(\frac{2+\sin 2 x}{1+\cos 2 x} e^x\)
Solution:
I = \(\int\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) e^x=\int\left(\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right) e^x\)
= \(\int\left(\frac{1+\sin x \cos x}{\cos ^2 x}\right) e^x\) = ∫(sec² x + tan x)e^x
Let f(x) = tan x ⇒ f'(x) = sec² x
∴ I = ∫[f(x) + f'(x)] e^x dx = e^xf(x) + C = e^x tan x + C

III.

Question 1.
Find the integral of \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Solution:
Given integral is \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\) dx
Let x = cosθ ⇒ dx = -sin θ dθ

Question 2.
Find the integral of \(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}\)
Solution:
Given integral of \(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}=\frac{\sqrt{x^2+1}}{x^4}\left[\log \left(x^2+1\right)-\log x^2\right]\)

Question 3.
Find the integral of (sin^-1 x)²
Solution:
Let I = ∫(sin^-1 x)² . 1 dx
Taking u = (sin^-1 x)², v = 1 and integrating by parts, we have