RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 7 Integrals Ex 7.6

Question 1.
x sin x
Answer:

= x( - cos x) - ∫[\(\frac{d}{d x}\) x ∫sin x dx] dx
= x (- cos x) - ∫ [1.(- cos x) dx]
= - x cos x + ∫ cos x dx
= - x cos x + sin x + C

Question 2.
x sin 3x
Answer:

Question 3.
x² e^x
Answer:

= x² e^x - 2[xe^x - ∫1.e^x dx]
= x²e^x - 2[xe^x - e^x] + C
= x²e^x - 2xe^x - 2e^x + C
= e^x (x² - 2x + 2) + C

Question 4.
x log x
Answer:

Question 5.
x log 2x
Answer:

Question 6.
x² log x
Answer:

Question 7.
x sin^-1 x
Answer:

Question 8.
x tan^-1 x
Answer:

Question 9.
x cos^-1 x
Answer:

Question 10.
(sin^-1 x)²
Answer:
Let I = ∫ (sin^-1 x)² dx
Putting sin^-1x = t
⇒ x = sin t
dx = cos t dt
∴ I = ∫ (sin^-1 x)² dx

I = t² ∫ cos t dt - ∫{\(\frac{d}{d t}\) (t²) ∫ cos t dt } dt
= t² sin t - ∫ 2t sin t dt
= t² sin t - 2 ∫ t sin t dt
I = t² sin t - 2I₁ ........ (1)

= t(- cos t) - ∫1.(- cos t) dt
= - t cos t + ∫cos t dt
= - t cos t + sin t + C₁
Putting the value of I₁ in equation (1), we have
I = t² sin t - 2(- t cos t + sin t + C₁)
= t² sin t + 2t cos t - 2 sin t - 2C₁
= t² sin t + 2t cos t - 2 sin t + C (∵ C = - 2C₁)
= t² sin t + 2t \(\sqrt{1-\sin ^{2} t}\) - 2 sin t + C
= x(sin^-1 x)² + 2 sin^-1 x\(\sqrt{1-x^{2}}\) - 2x + C

Question 11.
\(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\)
Answer:

= -[t sin t - ∫1.sin t dt]
= - t sin t + ∫ sin t dt
= - t sin t + (- cos t) + C
= - t sin t - cos t + C
= - t\(\sqrt{1-\cos ^{2} t}\) - cos t + C
= - (cos^-1 x)\(\sqrt{1-x^{2}}\) - x + C

Question 12.
x sec² x
Answer:

= x tan x - ∫1. tan x dx
= x tan x - ∫tan x dx
= x tan x - (- log |cos x|) + C
= x tan x + log |cos x| + C

Question 13.
tan^-1 x
Answer:

Question 14.
x (log x)²
Answer:

Question 15.
(x² + 1) log x
Answer:

Question 16.
e^x (sin x + cos x)
Answer:
Let I = ∫e^x (sin x + cos x) dx

= e^x - ∫ (cos x) e^x dx + ∫e^x cos x dx + C
= e^x - ∫e^x cos x dx + ∫e^x cos x dx + C
= e^x - ∫e^x cos x dx + ∫e^x cos x dx + C
= e^x sin x + C

Alternative: (i) ∫e^x (sin x + cos x) dx
Here f(x) = sin x and f’(x) = cos x
∵ ∫ e^x [f(x) + f’(x)] = e^x f(x) + C
∴ ∫e^x (sin x + cos x) dx = e^x sin x + C

(ii) Let I = ∫e^x (sin x + cos x) dx
Putting e^x sin x = t
⇒ (e^x cos x + e^x sin x)dx = dt
∴ I = ∫1 dt = t + C = e^x sin x + C

Question 17.
\(\frac{x e^{x}}{(1+x)^{2}}\)
Answer:

Question 18.
e^x \(\left(\frac{1+\sin x}{1+\cos x}\right)\)
Answer:

Question 19.
e^x \(\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Answer:

Alternative:
I = ∫e^x \(\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\) dx
Here f(x) = \(\frac{1}{x}\) and f'(x) = - \(\frac{1}{x^{2}}\)
From formula
∫e^x[f(x) + f'(x)] dx = e^x f(x) + C
I = \(\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x=e^{x} \frac{1}{x}+C=\frac{e^{x}}{x}+C\)

Question 20.
\(\frac{(x-3) e^{x}}{(x-1)^{3}}\)
Answer:

Alternative:

Question 21.
e^2x sin x
Answer:

I = e^2x ∫sin x dx - ∫ {\(\frac{d}{d x}\) e^2x ∫ sin x dx } dx
= e^2x (- cos x) - ∫2e^2x (- cos x) dx
= - e^2x cos x + 2∫e^2x cos x dx
I = - e^2x cos x + 2I₁ ........ (1)

⇒ I = e^2x ∫cos x dx - ∫ {\(\frac{d}{d x}\) e^2x ∫ cos x dx } dx
⇒ I = e^2x sin x - ∫ 2e^2x sin x dx + C₁
Putting the value of I₁ in equation (1), we get
I = - e^2x cos x + 2[e^2x sin x - 2 ∫e^2x sin x dx + C₁]
⇒ I = - e^2x cos x + 2e^2x sin x - 4 ∫e^2x sin x dx + 2C₁
⇒ I = - e^2x cos x + 2e^2x sin x - 4I + 2C₁
⇒ I + 4I = - e^2x cos x + 2e^2x sin x + 2C₁
⇒ 5I = - e^2x cos x + 2e^2x sin x + 2C₁
⇒ I = - \(\frac{1}{5}\)e^2x cos x + \(\frac{2}{5}\)e^2x sin x + \(\frac{2}{5}\)C₁
⇒ I = - \(\frac{1}{5}\)e^2x cos x + \(\frac{2}{5}\)e^2x sin x + C
(where C = \(\frac{2}{5}\)C₁)
I = \(\frac{e^{2 x}}{5}\) (2 sin x - cos x) + C

Question 22.
sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\)
Answer:

= 2θ tan θ - 2∫1.tan θ dθ
= 2θ tan θ - 2∫tan θ dθ
= 2θ tan θ - 2(- log | cos θ|) + C
= 2θ tan θ + 2 log |cos θ| + C

Question 23.
∫x² e^x3 dx equals:
(A) \(\frac{1}{3}\)e^x3 + C
(B) \(\frac{1}{3}\)e^x2 + C
(C) \(\frac{1}{2}\)e^x3 + C
(D) \(\frac{1}{2}\)e^x2 + C
Answer:
Let I = ∫x² e^x3 dx
Putting x³ = t
∴ I = \(\frac{1}{3}\)∫e^t dt = \(\frac{1}{3}\)e^t + C
= \(\frac{1}{3}\)e^x3 + C
Hence, (A) is the correct answer.

Question 24.
∫e^x sec x (1 + tan x) dx equals:
(A) e^x cos x + C
(B) e^x sec x + C
(C) e^x sin x + C
(D) e^x tan x + C
Answer:
I = ∫e^x sec x (1 + tan x) dx
= ∫e^x (sec x + sec x + tan x) dx
Here, f(x) = sec x and f'(x) = sec x tan x
Then, from formula
∫e^x [f(x) + f'(x)]dx = e^x f(x) + C
I = e^x sec x + C
Hence, (B) is the correct answer.