Evaluate: ∫ 5 x + 2 3 x - 4 · d x

∫ 5 x + 2 3 x − 4 ⋅ d x = ∫ 5 3 ( 3 x − 4 ) + 20 3 + 2 3 x − 4 d x = ∫ 5 3 ( 3 x − 4 ) + 26 3 3 x − 4 d x = ∫ [ 5 3 + ( 26 3 ) 3 x − 4 ] d x = 5 3 ∫ 1 d x + 26 3 ∫ 1 3 x − 4 d x = 5 x 3 + 26 3 ⋅ 1 3 log ⁡ | 3 x − 4 | + c = 5 x 3 + 26 3 log ⁡ | 3 x − 4 | + c .

Evaluate: ∫ x - 2 x + 5 · d x

∫ x − 2 x + 5 ⋅ dx = ∫ ( x + 5 ) − 7 x + 5 ⋅ dx ∫ ( x + 5 x + 5 − 7 x + 5 ) ⋅ dx = ∫ ( x + 5 ) 1 2 dx − 7 ∫ ( x + 5 ) − 1 2 dx = ( x + 5 ) 3 2 ( 3 2 ) − 7 ( x + 5 ) 1 2 ( 1 2 ) + c = 1 3 ( x + 5 ) 3 2 − 14 x + 5 + c

Evaluate: ∫ 2 x - 7 4 x - 1 · d x

Let I = ∫ 2 x − 7 4 x − 1 d x L e t 2 x − 7 = A 4 x - 1 + B ∴ 2 x − 7 = 4 A x + B - A B y E q u a t i n g t h e c o f f e c i e n t s o n b o t h s i d e s , w e g e t 4 A = 2 a n d B − A = − 7 ∴ A = 1 2 a n d B = − 7 + A = − 7 + 1 2 = − 13 2 ∴ 2 x − 7 = 1 2 4 x - 1 − 13 2 ∴ I = 1 2 ∫ ( 4 x − 1 ) − 13 4 x − 1 d x = 1 2 ∫ ( 4 x − 1 4 x − 1 − 13 4 x − 1 ) d x = 1 2 ∫ ( 4 x − 1 − 13 4 x − 1 ) d x = 1 2 ∫ ( 4 x − 1 ) 1 2 d x − 13 2 ∫ ( 4 x − 1 ) 1 2 d x = 1 2 [ ( 4 x − 1 ) 3 2 3 2 × 1 4 ] − 13 2 [ ( 4 x − 1 ) 1 2 1 2 × 1 4 ] ∴ I = 1 12 ( 4 x − 1 ) 3 2 − 13 4 4 x − 1 + c

Evaluate: ∫ sin 4 x cos 2 x · d x

Let I = ∫ sin 4 x cos 2 x · d x We know that sin 2 x = 2 sin x cos x So, sin 4 x = 2 sin 2 x cos 2 x ⇒ ∫ 2 sin 2 x cos 2 x cos 2 x · d x ⇒ ∫ 2 sin 2 x · d x ⇒ 2 . - cos 2 x 2 + C ∴ I = - cos 2 x + C

Evaluate: ∫ 1 + sin 5 x · d x

I = ∫ 1 + sin 5 x ⋅ d x = ∫ sin 2 x + cos 2 x + 2 sin 5 x 2 cos 5 x 2 ⋅ d x = ∫ ( sin 5 x 2 + cos 5 x 2 ) 2 ⋅ d x = ∫ ( sin 5 x 2 + cos 5 x 2 ) ⋅ d x = ∫ sin 5 x 2 d x + ∫ cos 5 x 2 ⋅ d x = 2 5 cos 5 x 2 − sin 5 x 2 + c

Evaluate ∫ cos 3 x d x

Let I = ∫ cos 3 x d x I = ∫ cos x × cos 2 x d x We know that sin 2 x + cos 2 x = 1 ⇒ I = ∫ cos x × 1 - sin 2 x d x Now, Put sin x = t ⇒ cos x d x = d t So, I = ∫ 1 - t 2 d t ⇒ I = t - t 3 3 + c ∵ ∫ x n d x = x n + 1 n + 1 + c ; n ≠ - 1 So, ⇒ I = sin x - sin 3 x + c Hence, ∫ cos 3 x d x = sin x - sin 3 x + c

Evaluate: ∫ 2 x - x + 3 · d x

∫ 2 x − x + 3 ⋅ d x = ∫ 2 x − x + 3 × x + x + 3 x + x + 3 ⋅ d x = ∫ 2 ( x + x + 3 ) x − ( x + 3 ) ⋅ d x = − 2 3 ∫ ( x + x + 3 ) ⋅ d x = − 2 3 ∫ x 1 2 d x − 2 3 ∫ ( x + 3 ) 1 2 ⋅ d x = − 2 3 ⋅ x 3 2 ( 3 2 ) − 2 3 ⋅ ( x + 3 ) 3 2 ( 3 2 ) + c = − 4 9 [ x 3 2 + ( x + 3 ) 3 2 ] + c .

Evaluate: ∫ 3 7 x - 2 - 7 x - 5 · d x

I = ∫ 3 7 x − 2 − 7 x − 5 ⋅ d x = ∫ 3 7 x − 2 − 7 x − 5 × 7 x − 2 + 7 x − 5 7 x − 2 + 7 x − 5 ⋅ d x = ∫ 3 ( 7 x − 2 + 7 x − 5 ) ( 7 x − 2 ) − ( 7 x − 5 ) ⋅ d x = ∫ ( 7 x − 2 + 7 x − 5 ) ⋅ d x = ∫ ( 7 x − 2 ) 1 2 ⋅ d x + ∫ ( 7 x − 5 ) 1 2 ⋅ d x = ( 7 x − 2 ) 3 2 3 2 × 1 7 + ( 7 x − 5 ) 3 2 3 2 × 1 7 + c = 2 21 ( 7 x − 2 ) 3 2 + 2 21 ( 7 x − 5 ) 3 2 + c .

E M B I B E

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 3 - Indefinite Integration>EXERCISE 3.1

Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 1: EXERCISE 3.1

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 1: EXERCISE 3.1

Attempt the practice questions on Chapter 3: Indefinite Integration, Exercise 1: EXERCISE 3.1 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 2 Standard 12 solutions are prepared by Experienced Embibe Experts.