Inter 2nd Year Maths Exercise 7f Solutions

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

Inter 2nd Year Maths Integrals Solutions Exercise 7f

Integrals Exercise 7f Solutions

I. Integrate the functions.

Question 1.
x sin x
Solution:
∫x sin x dx
By parts
∫ f(x) g(x) dx = f(x) ∫ g(x) dx – ∫ [f'(x) ∫ g(x) dx] dx
According to ILATE taking x as first function & second function sin x apply by parts
= x ∫ sin x dx – ∫ [1 ∫ (sin x dx)]dx
= -x cos x + ∫ cosx dx = -x cos x + sin x + C

Question 2.
x sin 3x
Solution:
Let I = ∫x sin 3x dx
According to ILATE taking x as first function & sin 3x as second function integrating by parts, we get
I = \(x \int \sin 3 x d x-\int\left[\left(\frac{d}{d x} x\right) \int \sin 3 x d x\right] d x\)
= \(\frac{-x \cos 3 x}{3}+\int \frac{\cos 3 x}{3} d x\) [∵ ∫ sin ax dx = \(\frac{-\cos a x}{a}\)]
⇒ I = \(\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x\) + C

Question 3.
x log x
Solution:
Let I = ∫x log x dx
According to ILATE taking log x as first function and x as second function & integrating by parts, we get,
I = \(\log x \int x d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x d x\right] d x\)
= \(\frac{x^2 \log x}{2}-\frac{1}{2} \int \frac{1}{x} x^2 d x=\frac{x^2 \log x}{2}-\frac{1}{2} \int x d x\)
= \(\frac{x^2 \log x}{2}-\frac{x^2}{4}\) + C

Question 4.
x log 2x
Solution:
Let I = ∫x log 2x dx
According to ILATE, taking log 2x as first function & x as second function and integrating by parts, we get,
I = \(\log 2 x \int x d x-\int\left[\left(\frac{d}{d x} \log 2 x\right) \int x d x\right] d x\)
= \(\frac{x^2}{2} \log 2 x-\int\left(\frac{1}{2 x} \times 2 \times \frac{x^2}{2}\right) d x\) = \(\frac{x^2 \log 2 x}{2}-\frac{1}{2} \int x d x\)
= \(\frac{x^2}{2} \log 2 x-\frac{x^2}{4}\) + C

Question 5.
x² log x
Solution:
Let I = ∫x² log x dx
According to ILATE. taking log x as first function and x² as second function and integrating by parts, we get
I = \(\log x \int x^2 d x-\int\left[\left(\frac{d}{d x} \log x\right) \int x^2 d x\right] d x\)
= \(\frac{x^3}{3} \log x-\int\left(\frac{1}{x} \cdot \frac{x^3}{3}\right) d x=\frac{x^3}{3} \log x-\frac{1}{3} \int x^2 d x\)
= \(\frac{x^3}{3} \log x-\frac{x^3}{9}\) + C

Question 6.
x sec² x
Solution:
Let I = ∫x sec² x dx
According to ILATE. taking x as first function and sec² x as second function and integrating by parts, we get,
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|sec x| + C
log|sec x| = log \(\left|\frac{1}{\cos x}\right|\) = log 1 – log|cos x|
= -log |cos x| (∵ log 1 = 0)
⇒ I = x tan x + log |cos x| + C

Question 7.
tan^-1 x
Solution:
Let I = ∫1.tan^-1 x dx
⇒ I = \(\tan ^{-1} x \int 1 d x-\int\left[\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 . d x\right] d x\)
= x tan^-1 x – ∫ \(\frac{1}{1+x^2}\) dx
Let 1 + x² = t ⇒ 2x = \(\frac{d t}{d x}\) ⇒ \(\frac{d t}{2 x}\) = dx
∴ I = x tan^-1 x – ∫ \(\frac{x}{t} \cdot \frac{d t}{2 x}\) = x tan^-1 x – \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt
= x tan^-1x – \(\frac{1}{2}\) log |t| + C
= x tan^-1x – \(\frac{1}{2}\) log |1 + x²| + C

Question 8.
(x² + 1)log x
Solution:
Let I = ∫(x² + 1) log x dx
According to ILATE, taking log x as first function & (x² + 1) as second function and integrating by parts we get
I = log x ∫ (x² + 1) dx – \(\int\left[\frac{d}{d x}(\log x) \int\left(x^2+1\right) d x\right] d x\)
⇒ I = log x (\(\frac{x^3}{3}\) + x) – ∫ \(\frac{1}{x}\)(\(\frac{x^3}{3}\) + x) dx
= (\(\frac{x^3}{3}\) + x) log x – ∫ (\(\frac{x^3}{3}\) + 1) dx
= (\(\frac{x^3}{3}\) + x) log x – \(\frac{x^3}{9}\) – x + C

Question 9.
e^x(sinx + cosx)
Solution:
Let I = ∫ e^x(sinx + cosx) dx
Let f(x) = sin x ⇒ f'(x) = cos x then,
I = ∫ e^x [f(x) + f'(x)] dx
We know that ∫ e^x [f(x) + f'(x)] dx = e^x dx
∴ I = e^xsin x + C

Question 10.
\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)
Solution:
Let I = ∫e^x\(\left(\frac{1}{x}-\frac{1}{x^2}\right)\)dx
Put f(x) = \(\frac{1}{x}\) ⇒ f'(x) = –\(\frac{1}{x^2}\)
∫ e^x [f(x) + f'(x)] dx = e^x f(x)
∴ I = \(\frac{e^x}{x}\) + C

II. Integrate the following functions.

Question 1.
x²e^x
Solution:
Let I = ∫ x²e^x dx
According to ILATE, taking x² as first function & e^x as second function and integrating by parts, we get
I = \(x^2 \int e^x d x-\int\left[\left(\frac{d}{d x} x^2\right) \int e^x d x\right] d x=x^2 e^x-\int 2 x e^x d x=x^2 e^x-2 \int x e^x d x\)
= x²e^x – ∫ [2xe^x dx]
Again, integrating by parts, we get
I = \(x^2 e^x-\left\{2 x \int e^x d x-2 \int\left(\frac{d}{d x}(x) \int e^x d x\right) d x\right\}\)
= x²e^x – 2xe^x + 2 ∫ e^x dx
= x²e^x – 2xe^x + 2e^x + C
∴ I = e^x(x² – 2x + 2) + C

Question 2.
x sin^-1 x
Solution:
Let I = ∫ xsin^-1x dx
According to ILATE, taking sin^-1 x as first function & x as second function and integrating by parts, we get,

Question 3.
x tan^-1 x
Solution:
Let I = ∫ x tan^-1 xdx
According to ILATE, taking tan^-1 x as first [unction & x as second function and integrating by parts, we get,

Question 4.
x cos^-1 x
Solution:
Let I = ∫ x cos^-1 xdx
Put cos^-1 x = t ⇒ x = cos t ⇒ dx = – sin t dt
I = ∫ x cos^-1 x dx = – ∫ t cos t . sin t dt
= –\(\frac{1}{2}\) ∫ t . 2 sin t cos t dt = – \(\frac{1}{2}\) ∫ t sin 2t dt
[∵ 2 sin x cos x = sin 2x]
= \(\frac{1}{4}\) t cos 2t – \(\frac{1}{4} \frac{\sin 2 t}{2}\) + C
= \(\frac{1}{4}\) t cos 2t – \(\frac{1}{8}\)sin 2t + C
= \(\frac{1}{4}\) t (2cot² t – 1) – \(\frac{1}{8}\)(2 sin t cos t) + C
[∵ cos 2x = 2 cos² x – 1 & sin 2x = 2 sin x cos x]
= \(\frac{1}{4}\) t (2cos²t – 1) – \(\frac{1}{4}\) (1 – cos² t)^1/2cos t + C
[∵ sin x = \(\sqrt{1-\cos ^2 x}\)]
∴ ∫ x cos^-1x dx = \(\frac{1}{4}\) cos^-1x(2x² – 1) – \(\frac{1}{4}\)(1 – x²)^1/2 x + C
[put cos^-1 x = t and cos t = x]
= \(\frac{1}{4}\)(2x² – 1)cos^-1x – \(\frac{x}{4} \sqrt{1-x^2}\) + C

Question 5.
\(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x \Rightarrow I=\int \cos ^{-1} x \frac{x}{\sqrt{1-x^2}} d x\)
According to ILATE consider cos^-1 x as first function & \(\frac{x}{\sqrt{1-x^2}}\) as second function & integrating by parts, we get

Question 6.
x(log x)²
Solution:
Let I = ∫ x(log x)² dx
According to ILATE taking (log x)² as first function & x as second function & integrating by parts, we get

Question 7.
\(\frac{x e^x}{(1+x)^2}\)
Solution:
Let I = ∫\(\frac{x e^x}{(1+x)^2}\) dx = ∫\(\mathrm{e}^{\mathrm{x}} \frac{(\mathrm{x}+1-1)}{(1+\mathrm{x})^2}\) dx
∫e^x\(\left[\frac{1}{(1+x)}-\frac{1}{(1+x)^2}\right]\) dx
Let f(x) = \(\frac{1}{(1+x)}\) ⇒ f'(x) = –\(\frac{1}{1+x)^2}\)
We know that ∫e^x [f(x) + f'(x)] dx = e^xf(x)
⇒ I = ∫e^x\(\left\{\frac{1}{1+x}-\frac{1}{(1+x)^2}\right\}\) dx
= \(\frac{e^x}{1+x}\) + C

Question 8.
\(e^x\left(\frac{1+\sin x}{1+\cos x}\right)\)
Solution:
∫e^x\(\left(\frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x}\right)\) dx
= ∫e^x\(\left(\frac{1}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right)\) dx
[∵ cos 2x = 2 cos²x – 1 & sin 2x = 2 sin x cos x]
= ∫e^x\(\left(\frac{\sec ^2 \frac{x}{2}}{2}+\tan \frac{x}{2}\right)\)dx = ∫e^x\(\left.\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right)\)dx
Let f(x) = tan\(\frac{x}{2}\) ⇒ f'(x) = \(\frac{\sec ^2 \frac{x}{2}}{2}\)
∴ ∫e^x\(\left(\frac{1+\sin x}{1+\cos x}\right)\) dx = e^x tan\(\frac{x}{2}\) + C
[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x)]

Question 9.
\(\frac{(x-3) e^x}{(x-1)^3}\)
Solution:
Let I = \(\int e^x\left[\frac{x-3}{(x-1)^3}\right] d x=\int e^x\left[\frac{x-1-2}{(x-1)^3}\right]\) dx
= \(\int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}-1}{(\mathrm{x}-1)^3}-\frac{2}{(\mathrm{x}-1)^3}\right) \mathrm{dx}=\int \mathrm{e}^{\mathrm{x}}\left(\frac{1}{(\mathrm{x}-1)^2}-\frac{2}{(\mathrm{x}-1)^3}\right) \mathrm{dx}\)
Let f(x) = \(\frac{1}{(x-1)^2}\) ⇒ f'(x) = \(\frac{-2}{(x-1)^3}\)
∴ I = \(\frac{\mathrm{e}^x}{(\mathrm{x}-1)^2}\) + C
[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x)]

Question 10.
e^2x sin x
Solution:
Let I = e^2x sin x dx
According to ILATE, taking sin x as first function & e^2x as second function & integrating by parts, we get

According to ILATE, taking sin x as first function & e^2x as second function & integrating by parts, we get

Put value of I₁ in eq(1), we get

Question 11.
\(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Solution:

Question 12.
\(\frac{2+\sin 2 x}{1+\cos 2 x} e^x\)
Solution:
Let I = ∫\(\frac{2+\sin 2 x}{1+\cos 2 x} e^x\) dx
= ∫\(\left(\frac{2}{1+\cos 2 x}+\frac{\sin 2 x}{1+\cos 2 x}\right)\) e^x dx
= ∫\(\left(\frac{2}{2 \cos ^2 x}+\frac{2 \sin x \cos x}{2 \cos ^2 x}\right)\) e^x dx
= ∫\(\left(\frac{1}{\cos ^2 x}+\tan x\right)\) e^x dx
= ∫(sec² + tan x)e^x dx = e^xtan x + C
[∵ ∫ e^x[f(x) + f'(x)] dx = e^xf(x)]

III. Integrate the functions

Question 1.
\(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Solution:

Question 2.
\(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}\)
Solution:

Question 3.
(sin^-1 x)²
Solution:
I = ∫(sin^-1 x)² dx
Put sin^-1 x = θ ⇒ x = sin θ ⇒ dx = cos θ dθ
= ∫θ²cos θ dθ
= ∫θ² cos θ dθ – ∫2θ (∫cosθ dθ) dθ
= θ² – 2 ∫θ sinθ dθ
= θ² sin θ – 2(θ ∫ sinθ dθ – ∫1(∫sinθ dθ)) dθ
= θ²sin θ – 2(-θ cosθ + ∫cosθ dθ)
= θ²sinθ + 2θ cosθ – 2 sinθ + C
= (sin^-1x)²x + 2sin^-1x\(\sqrt{1-\sin ^2 \theta}\) – 2x + C
= x(sin^-1x)² + 2\(\sqrt{1-x^2}\) sin^-1x – 2x + C