Inter 2nd Year Maths Exercise 7b Solutions

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7b will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7b

Integrals Exercise 7b Solutions

I. Integrate the following functions.

Question 1.
Find integral of \(\frac{2 x}{1+x^2}\)
Solution:
Let I = ∫\(\frac{2 x}{1+x^2}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
Put 1 + x² = t
Differentiating w.r.t ‘x’, we get 2x dx = dt
∴ I = ∫\(\frac{1}{t}\) dt = log |t| + C
= log |1 + x²| + C = log(1 + x²) + C

Question 2.
Find integral of \(\frac{(\log x)^2}{x}\)
Solution:
Let I = ∫\(\frac{(\log x)^2}{x}\) dx [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
Put log x = t
Differentiating w.r.t.x, we get \(\frac{1}{x}=\frac{d t}{d x}\) ⇒ dx = xdt
∴ I = ∫t² dt = \(\frac{t^3}{3}\) + C = \(\frac{(\log x)^3}{3}\) + C

Question 3.
Find integral of \(\frac{1}{x+x \log x}\)
Solution:
Let I = ∫\(\frac{1}{x+x \log x}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
Put 1 + log x = t
Differentiating w.r.t.x, we get \(\frac{1}{x}=\frac{d t}{d x}\) ⇒ \(\frac{1}{x}\) dx = xdt
∴ I = ∫\(\frac{1}{t}\) dt = log |t| + C
= log |1 + log x| + C

Question 4.
Find integral of sin x sin(cos x)
Solution:
Let I = ∫sin x sin(cos x)
Put cos x = t [∵ ∫ sin x dx = -cos x + C]
On differentiating w.r.t.x, we get
-sin x = \(\frac{d t}{d x}\) ⇒ dx = \(\frac{\mathrm{dt}}{-\sin x}\)
∴ I = ∫ sin x sin(t) = \(\frac{\mathrm{dt}}{-\sin x}\) = -∫sin t dt
= -(-cos t) + C = cos(cos x) + C

Question 5.
Find integral of sin(ax + b) cos(ax + b)
Solution:
Let I = ∫sin(ax + b) cos(ax + b) dx
= \(\int \frac{\sin 2(a x+b)}{2} d x\) [∵ ∫ sin x dx = -cos x + C]
= \(\frac{1}{2}\left[\frac{-\cos 2(a x+b)}{2 a}\right]+\) + C
= \(\frac{-1}{4 a}\)cos2(ax + b) + C.

Question 6.
Find integral of \(\sqrt{a x}+b\)
Solution:
\(\sqrt{a x}+b\) dx = ∫ (ax + b)^1/2 dx
= \(\frac{(a x+b)^{\frac{1}{2}+1}}{a\left(\frac{1}{2}+1\right)}\) + C [∵ ∫ (ax + b)ⁿ dx = \(\frac{(a x+b)^{n+1}}{a(n+1)}\)]
= \(\frac{(a x+b)^{3 / 2}}{a\left(\frac{3}{2}\right)}\) + C
= \(\frac{2}{3 a}\) (ax + b)^3/2 + C

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

Question 8.
Find integral of \(x \sqrt{1+2 x^2}\)
Solution:
Let I = ∫\(x \sqrt{1+2 x^2}\) dx
Put 1 + 2x² = t
On differentiating w.r.t. x, we get
4x = \(\frac{d t}{d x}\) ⇒ dx = \(\frac{d t}{d x}\)
∴ I = \(\int \mathrm{x} \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{4 \mathrm{x}}=\frac{1}{4} \int \sqrt{\mathrm{t}} \mathrm{dt}=\frac{1}{4} \int \mathrm{t}^{1 / 2} \mathrm{dt}\)
= \(\frac{1}{4} \frac{t^{(1 / 2)+1}}{(1 / 2)+1}\) + C = \(\frac{1}{4} \cdot \frac{2}{3} \cdot t^{3 / 2}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{1}{6}\) (1 + 2x²)^3/2 + C

Question 9.
Find integral of (4x + 2)\(\sqrt{x^2+x}+1\)
Solution:
Put I = ∫(4x + 2)\(\sqrt{x^2+x}+1\) dx
Let x² + x + 1 = t
On differentiating w.r.t.x, we get
2x + 1 =\(\frac{d t}{d x}\) ⇒ dx = \(\frac{\mathrm{dt}}{(2 \mathrm{x}+1)}\)
∴ I = \(\int(4 \mathrm{x}+2) \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{(2 \mathrm{x}+1)}\) [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\int 2(2 x+1) \sqrt{t} \frac{d t}{(2 x+1)}\) = \(2 \int \sqrt{t} d t\)
= \(2 \frac{t^{(1 / 2)+1}}{(1 / 2)+1}\) + C
= \(\frac{4}{3}\)(x² + x + 1)^3/2 + C

Question 10.
Find integral of \(\frac{1}{x-\sqrt{x}}\)
Solution:
\(\int \frac{1}{x-\sqrt{x}} d x=\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x\)
Put \(\sqrt{x}-1\) = t
⇒ Differentiating w.r.t. x, we get
\(\frac{1}{2 \sqrt{x}}=\frac{d t}{d x}\) ⇒ dx = \({2 \sqrt{x}}\) dt [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
∴ \(\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x\) = \(\int \frac{1}{\sqrt{x} t} 2 \sqrt{x} d t\)
= \(\int \frac{2}{t} d t\) = 2 . log|t| + C
= \(2 \log |\sqrt{x}-1|+C\)

Question 11.
Find integral of \(\frac{x}{\sqrt{x+4}}\), x > 0
Solution:

Question 12.
Find integral of (x³ – 1)^1/3x⁵
Solution:
I = ∫(x³ – 1)^1/3x⁵ dx = ∫(x³ – 1)^1/3x³x² dx
Let x³ – 1 = t ⇒ x³ = t + 1
Differentiating w.r.t. x, we get
= \(\frac{1}{3}\)∫(x³ – 1)^1/3x³(3x² dx) …..(i) [∵ \(\frac{d}{dx}\)(x³ – 1) = 3x²]
3x² = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{3 x^2}\)
∴ ∫ (x³ – 1)^1/3 x³ . x² dx
∫ t^1/3 (t + 1)x² \(\frac{d t}{3 x^2}\)

Question 13.
Find integral of \(\frac{x^2}{\left(2+3 x^3\right)^3}\)
Solution:

Question 14.
Find integral of \(\frac{1}{x(\log x)^m}\), x > 0, m ≠ 1
Solution:
\(\int \frac{1}{x(\log x)^m} d x\)
Let log x = t ⇒ \(\) ⇒ dx = xdt
∴ \(\int \frac{1}{x(\log x)^m} d x\) = \(\int \frac{1}{\mathrm{x}(\mathrm{t})^{\mathrm{m}}}\) x dt = ∫ t^-m dt
= \(\frac{\mathrm{t}^{-\mathrm{m}+1}}{-\mathrm{m}+1}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{(\log x)^{1-m}}{1-m}\) + C

Question 15.
Find integral of \(\frac{x}{9-4 x^2}\)
Solution:
\(\int \frac{x}{9-4 x^2} d x\)
Let 9 – 4x² = t ⇒ -8x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-8 x}\)
∴ \(\int \frac{x}{9-4 x^2} d x\) = \(\int \frac{x}{t} \frac{d t}{-8 x}\)
= \(\frac{1}{-8} \int \frac{1}{t} d t\)
= \(\frac{1}{-8}\) log|t| + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= \(\frac{1}{-8}\) log|9 – 4x²| + C

Question 16.
Find integral of e^2x+3 dx
Solution:
∫e^2x+3 dx
Let 2x + 3 = t ⇒ 2 = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{2}\)
∴ ∫e^(2x+3) dx = \(\int e^t \frac{d t}{2}\)
= \(\frac{1}{2}\)∫e^t dt
= \(\frac{1}{2}\)(e^t) + C [∵ ∫ e^x dx = e^x + C]
= \(\frac{1}{2}\)e^(2x+3) + C

Question 17.
Find integral of \(\frac{x}{e^{x^2}}\)
Solution:
\(\int \frac{x}{e^{x^2}} d x\)
Let x² = t ⇒ 2x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\mathrm{dt}}{2 \mathrm{x}}\)
∴ \(\int \frac{x}{e^{x^2}} d x=\int \frac{x}{e^t} \frac{d t}{2 x}\)
= \(\frac{1}{2}\) ∫e^-t dt = \(\frac{-1}{2}\) e^-t + C [∵ ∫ e^x dx = e^x + C]
= –\(\frac{1}{2}\) e^-x2 + C

Question 18.
Find integral of \(\frac{e^{\tan -x}}{1+x^2}\)
Solution:
\(\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x\)
Let tan^-1 x = t ⇒ \(\frac{1}{1+x^2}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = (1 + x²)dt
∴ \(\int \frac{e^{\tan ^{-1} x}}{1+x^2}\) dx = \(\int \frac{e^t}{1+x^2}\left(1+x^2\right) d t\)
= ∫e^t dt = e^t + C [∵ ∫ e^x dx = e^x + C]
= e^tan-1x + C

Question 19.
Find integral of \(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution:
\(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\) = \(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\)
= \(\int \frac{e^x\left(e^x-\frac{1}{e^x}\right)}{e^x\left(e^x+\frac{1}{e^x}\right)} d x\) = \(\int \frac{e^x-e^{-x}}{\left(e^x+e^{-x}\right)} d x\)
Let e^x + e^-x = t ⇒ e^x – e^-x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{e^x-e^{-x}}\)
∴ \(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\) =\( \int \frac{e^x-e^{-x}}{t} \cdot \frac{d t}{e^x-e^{-x}}\)
= \(\int \frac{1}{\mathrm{t}} \mathrm{dt}\)
= log|t| + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= log |e^x + e^-x| + C

Question 20.
Find integral of \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution:
\(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\)
Let e^2x + e^-2x = t [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
⇒ (2e^2x – 2e^-2x) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)
∴ \(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\) = \(\int \frac{e^{2 x}-e^{-2 x}}{t} \frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)
= \(\frac{1}{2} \int \frac{1}{t} d t\) = \(\frac{1}{2}\) log|t| + C
= \(\frac{1}{2}\) log|e^2x + e^-2x| + C

Question 21.
Find integral of tan²(2x – 3)
Solution:
∫ tan²(2x – 3) = ∫ sec²(2x – 3) dx – ∫ 1 dx [∵ tan²x = sec²x – 1]
Put 2x – 3 = t ⇒ 2 dx = dt ⇒ dx = \(\frac{1}{2}\) dt
∴ ∫ tan²(2x – 3)dx = ∫ sec² (2x – 3)dx – ∫ dx
= \(\frac{1}{2}\) ∫ sec²t dt – ∫ 1 dx
= \(\frac{1}{2}\) tan t – x + C [∵ ∫ sec²x dx = tan x + C; ∫ dx = x + C]
= \(\frac{1}{2}\) tan(2x – 3) – x + C

Question 22.
Find integral of sec²(7 – 4x)
Solution:
∫ sec²(7 – 4x) dx
Let 7 – 4x = t ⇒ -4 = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\mathrm{dt}}{-4}\)
∴ ∫sec²(7 – 4x)dx = ∫sec² t \(\frac{\mathrm{dt}}{-4}\)
= \(\frac{-1}{4}\)(tan t) + C [∵ ∫ sec²x dx = tan x + 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:

Question 26.
Find integral of \(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution:
\(\int \frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
Let \(\sqrt{x}\) = t ⇒ \(\frac{1}{2 \sqrt{x}}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(2 \sqrt{x}\) dt
∴ \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x=\int \frac{\cos t}{\sqrt{x}} 2 \sqrt{x} d t\)
= 2∫cos t dt = 2 sin t + C [∵ ∫ cos x dx = sin x + C]
= 2 sin\(\sqrt{x}\) + C

Question 27.
Find integral of \(\sqrt{\sin 2 x} \cos 2 x\)
Solution:
\(\int \sqrt{\sin 2 x} \cos 2 x d x\)
Let sin 2x = t ⇒ 2 cos 2x dx = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\mathrm{dt}}{2 \cos 2 \mathrm{x}}\)
∴ \(\int \sqrt{\sin 2 x} \cos 2 x d x\) = \(\int \sqrt{t} \cos 2 x \frac{d t}{2 \cos 2 x}\)
= \(\frac{1}{2} \int \sqrt{t} d t\)
= \(\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+C\) = \(\frac{1}{3}\)t^3/2 + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{1}{3}\) (sin 2x)^3/2 + C

Question 28.
Find integral of \(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution:
\(\int \frac{\cos x}{\sqrt{1+\sin x}} d x\) = ∫ (1 + sin x)^-1/2 cos x dx
Let 1 + sin x = t ⇒ cos x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{\cos x}\)
∴ ∫ (1 + sin x)^-1/2 cos x dx = ∫ (t)^-1/2 cos x \(\frac{d t}{\cos x}\)
= \(\frac{t^{-1 / 2+1}}{\left(-\frac{1}{2}+1\right)}+C\) [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= 2t^1/2 + C = \(2 \sqrt{1+\sin x}+C\)

Question 29.
Find integral of cot x log sin x
Solution:
∫cot x log sin x dx
Let log sin x = t ⇒ \(\frac{1}{\sin x}\) cos x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
⇒ cot x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{\cos x}\)
∴ ∫cot x log sin x dx = ∫ cot x . t \(\frac{d t}{\cos x}\)
= ∫ t dt = \(\frac{t^2}{2}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{(\log \sin x)^2}{2}+C\)

Question 30.
Find integral of \(\frac{\sin x}{1+\cos x}\)
Solution:
Let I = ∫\(\frac{\sin x}{1+\cos x}\) dx
Put 1 + cosx = t
⇒ -sin x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-\sin x}\)
∴ I = \(\int \frac{\sin x}{t} \times \frac{d t}{-\sin x}=-\int \frac{1}{t} d t\)
= – log |t | + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= -log|1 + cos x| + C

Question 31.
Find integral of \(\frac{\sin x}{(1+\cos x)^2}\)
Solution:
Let I = ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
Put 1 + cosx = t ⇒ -sin x = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{d t}{-\sin x}\)
∴ I = \(\int \frac{\sin x}{(1+\cos x)^2} d x=\int \frac{\sin x}{t^2} \times \frac{d t}{-\sin x}=-\int \frac{1}{t^2} \cdot d t\)
= -∫ t^-2 dt [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{-t^{-2+1}}{-2+1}+C\) = \(\frac{1}{t}\) + C
= \(\frac{1}{1+\cos x}\) + C

Question 32.
Find integral of \(\frac{(1+\log x)^2}{x}\)
Solution:
∫ \(\frac{(1+\log x)^2}{x}\) dx
Let 1 + log x = t ⇒ \(\frac{1}{x}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = x dt
∴ \(\int \frac{(1+\log x)^2}{x} d x=\int \frac{t^2}{x} x d t=\int t^2 d t\)
= \(\frac{t^3}{3}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + 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:
Let I = ∫ \(\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}\) dx
Put tan^-1 x⁴ = t ⇒ \(\frac{1}{1+x^8}\) . 4x³ = \(\frac{\mathrm{dt}}{\mathrm{dx}}\) ⇒ dx = \(\frac{\left(1+x^8\right)}{4 x^3}\) dt
∴ I = \(\int \frac{x^3 \sin t}{\left(1+x^8\right)} \cdot \frac{1+x^8}{4 x^3} d t=\frac{1}{4} \int \sin t d t\)
= –\(\frac{1}{4}\) cos t + C [∵ ∫ sin x dx = -cos x + C]
= –\(\frac{1}{4}\) cos x (tan^-1 x⁴ + C)

Question 35.
Find integral of \(\frac{x^3}{\sqrt{1-x^8}}\)
Solution:
∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx = ∫ \(\frac{x^3}{\sqrt{1-\left(x^4\right)^2}}\) dx
Put x⁴ = t ⇒ 4x³dx = dt ⇒ x³ dx = \(\frac{1}{4}\) dt
∴ \(\int \frac{1 / 4 d t}{\sqrt{1-t^2}}=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^2}}\)
= \(\frac{1}{4}\) sin^-1 t + C [∵ ∫ \(\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^2}}\) = sin^-1x + C]
= \(\frac{1}{4}\) sin^-1 (x⁴) + C

Question 36.
Find integral of cos³x e^{log sin x}
Solution:
Let I = ∫ cos³x e^{log sin x} dx
cos³ xe^logsinx = cos³ x sin x
Let cos x = t ⇒ -sin x dx = dt
∴ I = -∫ t³ dt = –\(\frac{t^4}{4}\) + C [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= –\(\frac{\cos ^4 x}{4}\) + C

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

Question 38.
Find integral of f'(ax + b)[f(ax + b)]ⁿ
Solution:
Let I = ∫ f'(ax + b)[f(ax + b)]ⁿ dx
Put f(ax + b) = t ⇒ af'(ax + b) dx = dt
⇒ dx = \(\frac{\mathrm{dt}}{\mathrm{af}^{\prime}(\mathrm{ax}+\mathrm{b})}\)
∴ I = ∫ f'(ax + b)tⁿ\(\frac{\mathrm{dt}}{\mathrm{af}^{\prime}(\mathrm{ax}+\mathrm{b})}\)
= \(\frac{1}{a} \int t^n d t\)
= \(\frac{1}{a}\left(\frac{\mathrm{t}^{\mathrm{n}+1}}{\mathrm{n}+1}\right)+\mathrm{C}\) [∵ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]
= \(\frac{1}{a} \frac{[f(a x+b)]^{n+1}}{n+1}+C\)

II.

Question 1.
Find integral of \(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\)
Solution:

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: