Find the integral of 1 x 2 + a 2 with respect to x and hence find ∫ 1 x 2 - 6 x + 13 d x .

Let I = ∫ 1 x 2 + a 2 d x Substitute x = a tan θ ⇒ θ = tan - 1 x a Now, differentiating, then d x = a sec 2 θ · d θ Therefore, I = ∫ 1 a tan θ 2 + a 2 a sec 2 θ · d θ ⇒ I = ∫ 1 a 2 tan 2 θ + a 2 a sec 2 θ · d θ ⇒ I = ∫ 1 a 2 tan 2 θ + 1 a sec 2 θ · d θ ⇒ I = ∫ 1 a 2 sec 2 θ a sec 2 θ · d θ ⇒ I = 1 a ∫ 1 · d θ ⇒ I = 1 a θ + c ⇒ I = 1 a tan - 1 x a + c Now, I 1 = ∫ 1 x 2 - 6 x + 13 d x ⇒ I 1 = ∫ 1 x - 3 2 - 9 + 13 d x ⇒ I 1 = ∫ 1 x - 3 2 + 4 d x ⇒ I 1 = ∫ 1 x - 3 2 + 2 2 d x ⇒ I 1 = 1 2 tan - 1 x - 3 2 + c Hence, ∫ 1 x 2 - 6 x + 13 d x = 1 2 tan - 1 x - 3 2 + c .

Prove that ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x and hence evaluate ∫ 0 a x x + a - x d x .

To prove: ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x Let ( a - x ) = t ⇒ - d x = d t When x = 0 ⇒ t = a and x = a ⇒ t = 0 Therefore, ∫ 0 a f ( a - x ) d x = - ∫ a 0 f ( t ) d t = ∫ 0 a f ( t ) d t (Apply property ∫ b a f ( x ) d x = - ∫ a b f ( x ) d x ) ⇒ ∫ 0 a f ( a - x ) d x = ∫ 0 a f ( x ) d x (Apply property ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ) Hence, proved. Let I = ∫ 0 a x x + a - x d x → 1 ⇒ I = ∫ 0 a a - x a - x + a - a - x d x = ∫ 0 a a - x a - x + x d x → 2 (Apply ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x ) Adding 1 and 2 , we get: 2 I = ∫ 0 a a - x a - x + x d x + ∫ 0 a x x + a - x d x ⇒ 2 I = ∫ 0 a a - x + x a - x + x d x ⇒ 2 I = ∫ 0 a 1 · d x ⇒ 2 I = x 0 a ⇒ 2 I = a - 0 ⇒ 2 I = a ⇒ I = a 2 Hence, ∫ 0 a x x + a - x d x = a 2 .

Find the integral of 1 x 2 - a 2 with respect to x and hence evaluate ∫ 1 x 2 - 16 d x .

Let I = ∫ 1 x 2 - a 2 d x Consider 1 x 2 - a 2 = 1 x - a x + a = A x - a + B x + a ⇒ 1 = A x + a + B x - a If x = a , then 1 = A 2 a + B 0 ⇒ A = 1 2 a If x = - a , then 1 = A 0 + B - 2 a ⇒ B = - 1 2 a So, 1 x 2 - a 2 = 1 2 a x - a - 1 2 a x + a Now, ∫ 1 x 2 - a 2 d x = ∫ 1 2 a x - a d x - ∫ 1 2 a x + a d x = 1 2 a ∫ 1 x - a d x - ∫ 1 x + a d x = 1 2 a log x - a - log x + a + C ∴ ∫ 1 x 2 - a 2 d x = 1 2 a log x - a x + a + C Now, ∫ 1 x 2 - 16 d x = ∫ 1 x 2 - 4 2 d x ∴ ∫ 1 x 2 - 16 d x = 1 8 log x - 4 x + 4 + C

Prove that ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x and hence evaluate ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x .

To prove ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x Let ( a - x ) = t ⇒ - d x = d t When x = 0 ⇒ t = a When x = a ⇒ t = 0 Now, ∫ 0 a f ( a - x ) d x = - ∫ a 0 f ( t ) d t = ∫ 0 a f ( t ) d t (Apply property ∫ b a f ( x ) d x = - ∫ a b f ( x ) d x ) ⇒ ∫ 0 a f ( a - x ) d x = ∫ 0 a f ( x ) d x (Apply property ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ) Hence, proved. Now, we have to find ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x Let I = ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x I = ∫ 0 π 2 cos 5 π 2 - x sin 5 π 2 - x + cos 5 π 2 - x d x ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x = ∫ 0 π 2 sin 5 x cos 5 x + sin 5 x d x So, 2 I = ∫ 0 π 2 sin 5 x cos 5 x + sin 5 x d x + ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x = ∫ 0 π 2 sin 5 x + cos 5 x cos 5 x + sin 5 x d x = ∫ 0 π 2 d x ⇒ 2 I = [ x ] 0 π 2 = π 2 - 0 ⇒ I = π 4 Therefore, ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x = π 4 .

Find ∫ x - 3 e x x - 1 3 d x

Let I = ∫ x - 3 e x x - 1 3 d x ⇒ I = ∫ x - 1 - 2 e x x - 1 3 d x ⇒ I = ∫ e x x - 1 x - 1 3 - 2 x - 1 3 d x ⇒ I = ∫ e x 1 x - 1 2 - 2 x - 1 3 d x ⇒ I = e x 1 x - 1 2 + c , c = constant of integration ∵ ∫ e x f x + f ' x d x = e x f x + c

∫ 0 π 2 cos 2 x d x

Let I = ∫ 0 π 2 cos 2 x d x ⇒ I = ∫ 0 π 2 cos 2 x + 1 2 d x We know that ∫ a b k f x d x = k ∫ a b f x d x ⇒ I = 1 2 ∫ 0 π 2 cos 2 x + 1 d x We know that ∫ a b f x ± g x d x = ∫ a b f x d x ± ∫ a b g x d x ⇒ I = 1 2 ∫ 0 π 2 cos 2 x d x + ∫ 0 π 2 d x ⇒ I = 1 2 sin 2 x 2 + x 0 π 2 ⇒ I = 1 2 sinπ 2 + π 2 - 0 ⇒ I = 1 2 0 + π 2 ⇒ I = π 4

Find the integral of 1 x 2 + a 2 with respect to x and hence evaluate ∫ 1 x 2 + 2 x + 2 d x .

Let I = ∫ 1 x 2 + a 2 d x Substitute x = a tan θ ⇒ θ = tan - 1 x a Differentiating with respect to θ on both sides ⇒ d x = a sec 2 θ d θ ⇒ I = ∫ a sec 2 θ a 2 tan 2 θ + a 2 d θ ⇒ I = ∫ a sec 2 θ a 2 tan 2 θ + 1 d θ ⇒ I = 1 a ∫ sec 2 θ sec 2 θ d θ ∵ sec 2 x - tan 2 x = 1 ⇒ I = 1 a ∫ d θ ⇒ I = 1 a θ + c ⇒ I = 1 a tan - 1 x a + c Now, ∫ 1 x 2 + 2 x + 2 d x = ∫ 1 x 2 + 2 x + 1 + 1 d x ⇒ ∫ 1 x 2 + 2 x + 2 d x = ∫ 1 x + 1 2 + 1 2 d x ⇒ ∫ 1 x 2 + 2 x + 2 d x = tan - 1 x + 1 + c .

Prove that ∫ − a a f ( x ) d x = { 2 ∫ 0 a f ( x ) d x , if f ( x ) is an event function 0 , if f ( x ) is an odd function and hence evaluate ∫ − 1 1 sin 5 ⁡ x cos 4 ⁡ x d x .

Let I = ∫ − a a f ( x ) d x ⇒ I = ∫ − a a f ( x ) d x = ∫ − a 0 f ( x ) d x + ∫ 0 a f ( x ) d x = I 1 + I 2 In I 1 , put x = - t ⇒ d x = - d t When x = - a , t = a and when x = 0 , t = 0 Therefore, I 1 = ∫ a 0 f ( − t ) ( − d t ) = − ∫ a 0 f ( − t ) d t = ∫ 0 a f ( − t ) d t ∵ ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x = ∫ 0 a f ( − x ) d x ∵ ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ∴ ∫ − a a f ( x ) d x = ∫ 0 a f ( − x ) d x + ∫ 0 a f ( x ) d x If f is an even function f - x = f x Therefore, ∫ − a a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( x ) d x = 2 ∫ 0 a f ( x ) d x If f is an odd function f - x = - f x Therefore, ∫ − a a f ( x ) d x = - ∫ 0 a f ( x ) d x + ∫ 0 a f ( x ) d x = 0 Now, let I 3 = ∫ − 1 1 sin 5 ⁡ x cos 4 ⁡ x d x . Let f ( x ) = sin 5 ⁡ x cos 4 ⁡ x Then f ( − x ) = sin 5 ⁡ ( − x ) cos 4 ⁡ ( − x ) = − sin 5 ⁡ x cos 4 ⁡ x = − f ( x ) i.e. f is an odd function Hence, I 3 = 0 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2019

Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Karnataka Board-2018

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Karnataka Board-2018

Attempt the free practice questions on Chapter 7: Integrals, Exercise 1: Karnataka Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Karnataka Board-2018 with Hints & Solutions

HARD

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2018>Q 6

Find the integral of $\frac{1}{x^{2} + a^{2}}$ with respect to $x$ and hence find $\int \frac{1}{x^{2} - 6 x + 13} d x$ .

HARD

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2018>Q 7

Prove that $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ and hence evaluate $\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x$ .

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2019>Q 14

Find the integral of $\frac{1}{x^{2} - a^{2}}$ with respect to $x$ and hence evaluate $\int \frac{1}{x^{2} - 16} d x$ .

HARD

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2019>Q 15

Prove that $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ and hence evaluate $\int_{0}^{\frac{π}{2}} \frac{\cos^{5} x}{\sin^{5} x + \cos^{5} x} d x$ .

EASY

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2020>Q 19

Find $\int \frac{(x - 3) e^{x}}{{(x - 1)}^{3}} d x$

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2020>Q 20

$\int_{0}^{\frac{π}{2}} \cos^{2} x d x$

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2020>Q 21

Find the integral of $\frac{1}{x^{2} + a^{2}}$ with respect to $x$ and hence evaluate $\int \frac{1}{x^{2} + 2 x + 2} d x$ .

HARD

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Karnataka Board-2020>Q 22

Prove that $\int_{- a}^{a} f (x) d x = {\begin{cases} 2 \int_{0}^{a} f (x) d x, if f (x) is an event function \\ 0, if f (x) is an odd function \end{cases}$ and hence evaluate $\int_{- 1}^{1} \sin^{5} x \cos^{4} x d x$ .

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