AP Inter 2nd Year Maths Exercise 7f Solutions

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