RBSE Maths Class 12 Chapter 9: Integration Important Questions and Solutions

RBSE Class 12 Maths Chapter 9 – Integration Important questions and solutions are available here. The solutions for the important questions of this chapter have detailed explanations. The RBSE Class 12 important questions and solutions given at BYJU’S will help the students in better understanding of all the concepts and to score maximum marks in the exams.

Chapter 9 of RBSE Class 12 has seven exercises, all of which focus on integrating different functions of simple to complex. Each exercise possesses a different method for integrating the given function. After practicing these solutions, students will be able to integrate any type of function in the exam easily.

RBSE Maths Chapter 9: Exercise 9.1 Textbook Important Questions and Solutions

Question 1: Integrate the following function with respect to x.

Solution:

= ∫x² dx

= (x³/3) + C

Question 2: Evaluate: ∫ cot x (tan x – cosec x) dx

Solution:

∫ cot x (tan x – cosec x) dx

= ∫ cot x tan x dx – ∫ cot x cosec x dx

= ∫ 1 dx – ∫ cosec x cot x dx

= x + cosec x + C

RBSE Maths Chapter 9: Exercise 9.2 Textbook Important Questions and Solutions

Question 3: Integrate the function x sin x² with respect to x.

Solution:

Let x² = t

⇒ 2x dx = dt

⇒ x dx = dt/2

Now,

∫x sin x² dx = ∫sin x² . x dx

= ∫sin t (dt/2)

= (1/2) (-cos t) + C

= -(1/2) cos x² + C

Question 4: Integrate the following function with respect to x.

1/[x(1 + log x)]

Solution:

Let 1 + log x = t

(1/x) dx = dt

∫1/[x(1 + log x)] dx = ∫[1/(1 + log x)] (1/x) dx

= ∫(1/t) dt

= log |t| + C

= log |1 + log x| + C

RBSE Maths Chapter 9: Exercise 9.3 Textbook Important Questions and Solutions

Question 5: Integrate the following function with respect to x.

(50 + 2x²)

Solution:

∫1/(50 + 2x²) dx

= (1/2) ∫1/(25 + x²) dx

= (1/2) ∫1/(5² + x²) dx

= (1/2) ∫1/(x² + 5²) dx

= (1/2) × (1/5) tan^-1(x/5) + C

= (1/10) tan^-1(x/5) + C

Question 6: Integrate the following function with respect to x.

1/√(1 + 4x²)

Solution:

∫1/√(1 + 4x²) dx

= ∫1/√[1² + (2x)²] dx

= 1/2 log |2x + √[1 + (2x)²] | + C

= (1/2) log |2x + √(4x² + 1) | + C

RBSE Maths Chapter 9: Exercise 9.4 Textbook Important Questions and Solutions

Question 7: Integrate the following function with respect to x.

3x/(x + 1)(x – 2)

Solution:

∫[3x/(x + 1)(x – 2)] dx

Let 3x/(x + 1)(x – 2) = [A/(x + 1)] + [B/(x – 2)]

⇒ 3x = Ax – 2A + Bx + B

3x = (A + B)x – 2A + B

By comparing the coefficients,

A + B = 3 ….(i)

-2A + B = 0

B = 2A ….(ii)

From (i) and (ii),

A + 2A = 3

3A = 3

A = 3/3 = 1

B = 2A = 2(1) = 2

Thus,

3x/(x + 1)(x – 2) = [1/(x + 1)] + [2/(x – 2)]

∫[3x/(x + 1)(x – 2)] dx = ∫1/(x + 1) dx + 2∫1/(x – 2) dx

= log |x + 1| + 2 log |x – 2| + C

Question 8: Integrate the following function with respect to x.

(1 – cos x)/[cos x (1 + cos x)]

Solution:

∫(1 – cos x)/[cos x (1 + cos x)] dx

= ∫1/[cos x (1 + cos x)] dx – ∫cos x/[cos x (1 + cos x)] dx

= ∫(1/cos x) dx – ∫1/(1 + cos x) dx – ∫1/(1 + cos x) dx

= ∫(1/cos x) dx – ∫2[1/(1 + cos x)] dx

= ∫(1/cos x) dx – 2∫1/ [2 cos²(x/2)] dx

= ∫sec x dx – ∫sec²(x/2) dx

= log |sec x + tan x| – 2 tan(x/2) + C

RBSE Maths Chapter 9: Exercise 9.5 Textbook Important Questions and Solutions

Question 9: Integrate the following function with respect to x.

1/(x² + 2x + 10)

Solution:

∫1/(x² + 2x + 10) dx

= ∫1/(x² + 2 × 1 × x + 1² + 9) dx

= ∫1/[(x + 1)² + 3²] dx

= (1/3) tan^-1[(x + 1)/3) + C

Question 10: Integrate the following function with respect to x.

cos x/(sin²x + 4 sin x + 5)

Solution:

Let sin x = t

⇒ cos x dx = dt

∫cos x/(sin²x + 4 sin x + 5) dx

= ∫ dt/(t² + 4t + 5)

= ∫dt/[(t² + 2 × 2t + 2²) + 1]

= ∫dt/[(t + 2)² + 1]

= (1/1) tan^-1[(t + 2)/1] + C

= tan^-1(t + 2) + C

= tan^-1(sin x + 2) + C

RBSE Maths Chapter 9: Exercise 9.6 Textbook Important Questions and Solutions

Question 11: Integrate the following function with respect to x.

x³ sinx

Solution:

Question 12: Integrate the following function with respect to x.

cos √x

Solution:

Let √x = t

⇒ x = t²

⇒ dx = 2t dt

∫cos √x dx = ∫cos t. 2t dt

= 2∫t. cos t dt

= 2[t ∫cos t dt – ∫ {d/dt (t). ∫cos t dt} dt]

= 2[t sin t – ∫sin t dt]

= 2[t sin t + cos t] + C

= 2[√x sin √x + cos √x] + C

RBSE Maths Chapter 9: Exercise 9.7 Textbook Important Questions and Solutions

Question 13: Integrate the following function with respect to x.

sin (log x)

Solution:

Let I = ∫ sin (log x) dx

And let log x = t ⇒ x = e^t ⇒ dx = e^t dt

Now,

I = ∫sin t et dt

= -e^t cos t – ∫e^t (-cos t) dt

= -e^t cos t + ∫e^t cos t dt

= -e^t cos t + [e^t sin t – ∫e^t sin t dt]

= -e^t cos t + e^t sin t – I

⇒ I + I = e^t (sin t – cos t)

⇒ 2I = e^t (sin t – cos t)

⇒ I = (e^t/2)(sin t – cos t) + C

= (x/2)[sin (log x) – cos (log x)] + C

Question 14: Integrate the following function with respect to x.

x² √(a⁶ – x⁶)

Solution:

∫x² √(a⁶ – x⁶) dx = ∫√[(a³)² – (x³)²] x² dx

Let x³ = t

⇒ 3x² dx = dt

⇒ x² dx = dt/3

∫x² √(a⁶ – x⁶) dx =∫√[(a³)² – t²] (dt/3)

= (1/3) ∫√[(a³)² – t²] dt

= (1/3) {(1/2)t √[(a³)² – t²] + (1/2) (a³)² sin^-1(t/a³)} + C

= (1/3) {(1/2)t √(a⁶ – t²) + (a⁶/2) sin^-1(t/a³)} + C

= (1/3) [(x³/2) √(a⁶ – x⁶) + (a⁶/2) sin^-1(x³/a³)] + C

RBSE Maths Chapter 9: Additional Important Questions and Solutions

Question 1: ∫ (1 – cos 2x)/(1 + cos 2x) dx =

(a) tan x + x + C

(b) cot x + x + C

(c) tan x – x + C

(d) cot x – x + C

Solution:

Correct answer: (c)

∫ (1 – cos 2x)/(1 + cos 2x) dx

= ∫(1 – 1 + 2 sin²x)/ (1 + 2 cos²x – 1) dx

= ∫ sin²x/cos²x dx

= ∫tan²x dx

= ∫(sec²x – 1) dx

= ∫sec²x dx – ∫1 dx

= tan x – x + C