AP Inter 2nd Year Maths Exercise 7b Solutions

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

AP Inter 2nd Year Maths Integrals Solutions Exercise 7b

I.

Question 1.
Find integral of \(\frac{2 x}{1+x^2}\)
Solution:
Put 1 + x² = t ⇒ 2xdx = dt
∴ ∫\(\frac{2 x}{1+x^2}\) dx = ∫\(\frac{d t}{t}\) = ∫\(\frac{1}{t}\) dt = log |t| + c = log |1 + x²| + c = log(1 + x²) + c. [∵ t = 1 + x²]

Question 2.
Find integral of \(\frac{(\log x)^2}{x}\)
Solution:
Put log x = t ⇒ \(\frac{1}{x}\) dx = dt ⇒ \(\frac{dx}{x}\) = dt
∴ ∫\(\frac{(\log x)^2}{x}\) dx = ∫(log x)²\(\left(\frac{d x}{x}\right)\) = ∫t² dt = \(\frac{t^3}{3}\) + c = \(\frac{1}{3}\)(log x)³ + c [∵ t = log x]

Question 3.
Find integral of \(\frac{1}{x+x \log x}\)
Solution:
Put 1 + log x = t ⇒ \(\frac{d x}{x}\) = dx
∴ \(\int \frac{1}{x+x \log x} d x=\int \frac{1}{1+\log x}\left(\frac{d x}{x}\right)\) = ∫\(\frac{1}{t}\) dt = log |t| + c = log |1 + log x| + c [∵ t = 1 + log x]

Question 4.
Find integral of sinx sin(cos x)
Solution:
We have to find ∫sin x sin(cos x) dx = -∫sin(cos x)(-sin x)dx
Put cos x = t ⇒ -sin x dx = dt
∴ ∫sinx sin(cos x)dx = -∫sin(cox x)(-sin x dx)
= -∫sint dt = -(-cos t) + c = cos t + c = cos(cos x) + c

Question 5.
Find integral of sin(ax + b) ocs(ax + b)
Solution:
∫sin(ax + b) cos(ax + b) dx = \(\frac{1}{2}\)∫2sin(ax + b)cos(ax + b)dx
= \(\frac{1}{2}\)∫sin2(ax + b) dx = \(\frac{1}{2}\)∫sin(2ax + 2b) dx [∵ 2 sin A cos A = sin 2A]
= \(\frac{1}{2} \frac{[-\cos (2 a x+2 b)]}{2 a}\) + c = \(\frac{-1}{4 a}\)cos2(ax + b) + c. [∵ sin(ax + b) dx = \(-\frac{1}{a}\)cos(ax + b) + c]

Question 6.
Find integral of \(\sqrt{a x}+b\)
Solution:

Question 7.
Find integral of \(x \sqrt{x+2}\)
Solution:
∫\(x \sqrt{x+2}\) dx = ∫\(x \sqrt{x+2}\)dx = ∫((x + 2) – 2)\(\sqrt{x+2}\) dx

Question 8.
Find integral of \(x \sqrt{1+2 x^2}\)
Solution:
Let I = ∫\(x \sqrt{1+2 x^2}\) dx = \(\frac{1}{4}\)∫\(\sqrt{1+2 x^2}\)(4xdx) ………….(i) [∵ \(\frac{d}{d x}\)(1 + 2x²) = 0 + 2.2x = 4x]
Put 1 + 2x² = t ⇒ 4xdx = dt
∴ From (i), I = \(\frac{1}{4}\)∫\(\sqrt{t}\)dt = \(\frac{1}{4}\)∫t^1/2 dt
[∵ t = 1 + 2x²]

Question 9.
Find integral of (4x + 2)\(\sqrt{x^2+x}+1\)
Solution:
Let I = ∫(4x + 2)\(\sqrt{x^2+x}+1\) dx = ∫2(2x + 1)\(\sqrt{x^2+x+1} d x\)
= ∫2\(\sqrt{x^2+x+1}\)(2x + 1) dx ……(i)
Put x² + x + 1 = t ⇒ (2x + 1)dx = dt
From (i), I = ∫2\(\sqrt{t}\) dt = 2∫t^1/2 dt
= \(2 \frac{t^{3 / 2}}{\frac{3}{2}}+c=\frac{4}{3} t^{3 / 2}+c=\frac{4}{3}\left(x^2+x+1\right)^{3 / 2}+c\) [∵ t = x² + x + 1]

Question 10.
Find integral of \(\frac{1}{x-\sqrt{x}}\)
Solution:
Let I = \(\int \frac{1}{x-\sqrt{x}} d x\) …………….(i)
Put \(\sqrt{\text { Linear }}\) = t, i.e., \(\sqrt{x}\) = t ⇒ x = t² ⇒ dx = 2t dt
∴ From (i), I = \(\int \frac{1}{t^2-t}\)2tdt = 2∫\(\frac{t}{t(t-1)}\) dt
= 2\(\int \frac{1}{t-1}\) dt = 2log |t – 1| + c = 2log \(|\sqrt{x}-1|\) + c [∵ \(\int \frac{1}{a x+b} d x=\frac{1}{a}\) log |ax + b|]

Question 11.
Find integral of \(\frac{x}{\sqrt{x+4}}\), x > 0
Solution:
Let I = \(\int \frac{x}{\sqrt{x+4}} d x\) ………….(i)

Question 12.
Find integral of (x³ – 1)^1/3x⁵
Solution:
Let I = ∫(x³ – 1)^1/3x⁵ dx = ∫(x³ – 1)^1/3x³x² dx
= \(\frac{1}{3}\)∫(x³ – 1)^1/3x³(3x² dx) …..(i) [∵ \(\frac{d}{dx}\)(x³ – 1) = 3x²]
Put x³ – 1 = t ⇒ x³ = t + 1 ⇒ 3x² = \(\frac{dt}{dx}\) ⇒ 3x² dx = dt

Question 13.
Find integral of \(\frac{x^2}{\left(2+3 x^3\right)^3}\)
Solution:
Let I = \(\) …….(i) [\(\frac{d}{d x}\int \frac{x^2}{\left(2+3 x^3\right)^3} d x=\frac{1}{9} \int \frac{9 x^2}{\left(2+3 x^3\right)^3} d x\)(2 + 3x³) = 9x²]
Put 2 + 3x³ = t ⇒ 9x² dx = dt
∴ Fron (i), I = \(\frac{1}{9} \int t^{-3} d t=\frac{1}{9}\left(\frac{t^{-2}}{-2}\right)+c=\frac{-1}{18 t^2}+c=\frac{-1}{18\left(2+3 x^3\right)^2}+c\) [∵ t = 2 + 3x³]

Question 14.
Find integral of \(\frac{1}{x(\log x)^m}\), x > 0, m ≠ 1
Solution:
Let I = \(\int \frac{1}{x(\log x)^m} d x(x>0) \Rightarrow I=\int \frac{\frac{I}{x} d x}{(\log x)^m}\) ……(i)
Put log x = t ⇒ \(\frac{d x}{x}\) = dt
From (i), I = \(\int \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{m}}}=\int \mathrm{t}^{-\mathrm{m}} \mathrm{dt}=\frac{\mathrm{t}^{-\mathrm{m}+1}}{-\mathrm{m}+1}+\mathrm{c}\) (Assuming m ≠ 1)
= \(\frac{(\log x)^{1-m}}{1-m}\) + c [∵ t = log x]

Question 15.
Find integral of \(\frac{x}{9-4 x^2}\)
Solution:
Let I = \(\int \frac{x}{9-4 x^2} d x=\frac{-1}{8} \int \frac{-8 x}{9-4 x^2} d x\) ……….(i) [∵ \(\frac{d}{d x}\)(9 – 4x²) = -8x]
Put 9 – 4x² = t ⇒ -8xdx = dt [∵ \(\) \int \frac{f^{\prime}(x)}{f(x)} d x= logf(x) + c]

Question 16.
Find integral of e^2x+3
Solution:
Put 2x + 3 = t ⇒ 2 dx = dt [∵ ∫e^ax+bdx = \(\frac{1}{a}\)e^{ax + b} + c]
∴ ∫e[sup]2x+3[/sup]dx = \(\frac{1}{2}\)∫e^t dt = \(\frac{1}{2}\)(e^t) + C = \(\frac{1}{2}\)e^(2x+3) + C

Question 17.
Find integral of \(\frac{x}{e^{x^2}}\)
Solution:
Put x² = t ⇒ 2xdx = dt
∴ \(\int \frac{x}{e^{x^2}} d x=\frac{1}{2} \int \frac{1}{e^t} d t=\frac{1}{2} \int e^{-t} d t=\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+C=-\frac{1}{2} e^{-x^2}+C=\frac{-1}{2 e^{x^2}}+C\)

Question 18.
Find integral of \(\frac{e^{\tan -x}}{1+x^2}\)
Solution:
Put tan^-1 x = t ⇒ \(\frac{1}{1+x^2}\)dx = dt ∴ \(\int \frac{e^{\tan ^{-1} x}}{1+x^2}\) dx = ∫e^t dt = e^t + C = e^tan-1x + C

Question 19.
Find integral of \(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution:
\(\frac{e^{2 x}-1}{e^{2 x}+1}\) Dividing Nr. and Dr. by e^x we get \(\frac{\frac{e^{2 x}-1}{e^x}}{\frac{e^{2 x}+1}{e^x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\) [\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]
Put e^x + e^-x = t ⇒ (e^x – e^-x)dx = dt
⇒ \(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{e^x-e^{-x}}{e^x+e^{-x}} d x=\int \frac{d t}{t}\) = log|t| + C = log |e^x + e^-x| + C

Question 20.
Find integral of \(\)
Solution:
Put e^2x + e^-2x = t ⇒ (2e^2x – 2e^-2x) dx = dt ⇒ (2e^2x – 2e^-2x) dx = dt

Question 21.
Find integral of tan²(2x – 3)
Solution:
We have tan²(2x – 3) = sec²(2x – 3) – 1
Put 2x – 3 = t ⇒ 2 dx = dt
⇒ \(\int \tan ^2(2 x-3) d x=\int\left[\sec ^2(2 x-3)-1\right] d x\)
= \(\frac{1}{2} \int \sec ^2 \mathrm{tdt}-\int 1 \mathrm{dx}=\frac{1}{2} \tan \mathrm{t}-\mathrm{x}+\mathrm{C}=\frac{1}{2} \tan (2 \mathrm{x}-3)-\mathrm{x}+\mathrm{C}\)

Question 22.
Find integral of sec²(7 – 4x)
Solution:
Put 7 – 4x = t ⇒ -4 dx = dt
∴ ∫sec²(7 – 4x)dx = \(\frac{-1}{4}\)∫sec² tdt = \(\frac{-1}{4}\)(tan t) + C = \(\frac{-1}{4}\)tan(7 – 4x) + C

Question 23.
Find integral of \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\)
Solution:

Question 24.
Find integral of \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Solution:

Question 25.
Find integral of \(\frac{1}{\cos ^2 x(1-\tan x)^2}\)
Solution:
We have \(\frac{1}{\cos ^2 x(1-\tan x)^2}=\frac{\sec ^2 x}{(1-\tan x)^2}\)
Put (1 – tan x) = t ⇒ -sec² xdx = dt
∴ \(\int \frac{\sec ^2 x}{(1-\tan x)^2} d x=\int \frac{-d t}{t^2}=-\int t^{-2} d t=\frac{1}{t}+C=\frac{1}{(1-\tan x)}+C\)

Question 26.
Find integral of \(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution:
Put \(\sqrt{x}\) = t ⇒ \(\frac{1}{2 \sqrt{x}}\)dx = dt ⇒ \(\int \frac{\cos \sqrt{x}}{\sqrt{x}}\) = 2∫costdt = 2 sin t + C = 2 sin\(\sqrt{x}\) + C

Question 27.
Find integral of \(\sqrt{\sin 2 x} \cos 2 x\)
Solution:
Put sin 2x = t ⇒ 2 cos 2x dx = dt
∴ \(\int \sqrt{\sin 2 x} \cos 2 x d x=\frac{1}{2} \int \sqrt{t} d t=\frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C=\frac{1}{3} t^{\frac{3}{2}}+C=\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C\)

Question 28.
Find integral of \(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution:
Put sin 2x = t ⇒ cosxdx = dt
∴ \(\int \frac{\cos x}{\sqrt{1+\sin x}} d x=\int \frac{d t}{\sqrt{t}}=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C=2 \sqrt{t}+C=2 \sqrt{1+\sin x}+C\)

Question 29.
Find integral of cotx logsin x
Solution:
Put logsin x = t ⇒ \(\frac{1}{\sin x}\) cos xdx = dt ∴ cot x dx = dt
⇒ ∫cotx log sin xdx = ∫ tdt = \(\frac{t^2}{2}\) + C = \(\)(log sin x)² + \(\frac{1}{2}\)

Question 30.
Find integral of \(\frac{\sin x}{1+\cos x}\)
Solution:
Put 1 + cosx = t ⇒ -sinx dx = dt
⇒ ∫\(\frac{\sin x}{1+\cos x}\) dx = ∫\(-\frac{\mathrm{dt}}{\mathrm{t}}\) = – log |t | + C = -log|1 + cos x| + C

Question 31.
Find integral of \(\frac{\sin x}{(1+\cos x)^2}\)
Solution:
Put 1 + cosx = t ⇒ -sinx dx = dt
∴ \(\int \frac{\sin x}{(1+\cos x)^2} d x=\int-\frac{d t}{t^2}=-\int t^{-2} d t=\frac{1}{t}+C=\frac{1}{(1+\cos x)}+C\)

Question 32.
Find integral of \(\frac{(1+\log x)^2}{x}\)
Solution:
Put 1 + log x = t ⇒ \(\frac{1}{x}\)dx = dt ∴ \(\int \frac{(1+\log x)^2}{x} d x=\int t^2 d t=\frac{t^3}{3}+C=\frac{(1+\log x)^3}{3}+C\)

Question 33.
Find integral of \(\frac{(x+1)(x+\log x)^2}{x}\)
Solution:

Question 34.
Find integral of \(\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}\)
Solution:
Put x³ = t ⇒ 4x³dx = dt

Question 35.
Find integral of \(\frac{x^3}{\sqrt{1-x^8}}\)
Solution:
Put x³ = t ⇒ 4x³dx = dt
∴ \(\int \frac{x^3}{\sqrt{1-x^8}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^2}}=\frac{1}{4} \sin ^{-1} t+C=\frac{1}{4} \sin ^{-1}\left(x^4\right)+C\)

Question 36.
Find integral of cos³x e^{logsin x}
Solution:
cos³ xe^logsinx = cos³ x sin x
Let cos x = t ⇒ -sin xdx = dt
∴ \(\int \cos ^3 x e^{\log \sin x} d x=\int \cos ^3 x \sin x d x=-\int t^3 d t=-\frac{t^4}{4}+C=-\frac{\cos ^4 x}{4}+C\)

Question 37.
Find integral of e^{3log x}(x⁴ + 1)^-1
Solution:
e^{3log x}(x⁴ + 1)^-1 = e^{log x3}(x⁴ + 1)^-1 = \(\frac{x^3}{\left(x^4+1\right)}\) [∵ \(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]
Let x⁴ + 1 = t ⇒ 4x³ dx = dt
⇒ \(\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=\int \frac{x^3}{\left(x^4+1\right)} d x=\frac{1}{4} \int \frac{d t}{t}=\frac{1}{4} \log |t|+C=\frac{1}{4} \log \left|x^4+1\right|+C\)

Question 38.
Find integral of f'(ax + b)[f(ax + b)]ⁿ
Solution:
Given integral is f'(ax + b)[f(ax + b)]ⁿ
Put f(ax + b) = t ⇒ af'(ax + b) dx = dt
⇒ f'(ax + b)[f(ax + b)]ⁿ dx = \(\frac{1}{a} \int t^n d t=\frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]=\frac{1}{a(n+1)}(f(a x+b))^{n+1}+C\)

II.

Question 1.
Find integral of \(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\)
Solution:
Given integrand is \(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\). Multiplying and dividing by x^-3, we get

Question 2.
Find integral of \(\frac{1}{1+\cot x}\)
Solution:

Question 3.
Find integral of \(\frac{1}{1-\tan x}\)
Solution:

Question 4.
Find integral of \(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution: