Integrate the following functions w.r.t.x ∫ 1 4 x 2 - 3 · d x

I = ∫ 1 4 x 2 − 3 ⋅ d x = 1 4 ∫ 1 x 2 − 3 4 ⋅ d x = 1 4 ∫ 1 x 2 − ( 3 2 ) 2 ⋅ d x = 1 4 1 2 ( 3 2 ) log ⁡ | x − 3 2 x + 3 2 | + c = 1 4 3 log ⁡ | 2 x − 3 2 x + 3 | + c

Evaluate the following ∫ 1 2 x 2 - 5 · d x

I = ∫ 1 2 x 2 − 5 d x = 1 2 ∫ 1 x 2 − 5 2 d x = 1 2 ∫ 1 x 2 − ( 5 2 ) 2 d x = 1 2 log ⁡ x + x 2 − 5 2 + c .

Evaluate the following ∫ 9 + x 9 - x · d x

Let I = 9 + x 9 - x d x I = ∫ 9 + x 9 + x × 9 + x 9 - x d x = ∫ 9 + x 9 - x 2 d x I = 9 ∫ 1 9 2 - x 2 d x + ∫ x 9 2 - x 2 d x Put 9 2 - x 2 = t ⇒ x d x = - d t 2 to the second integral, we get I = ∫ 1 9 2 - x 2 d x - 1 2 ∫ d t t I = ∫ 1 9 2 - x 2 d x - 1 2 ∫ t - 1 2 + c I = sin - 1 x 9 - 1 2 t 1 2 1 2 + c I = sin - 1 x 9 - 9 2 - x 2 + c

Evaluate the following ∫ 10 + x 10 - x · d x

Let I = 10 + x 10 - x d x I = ∫ 10 + x 10 + x × 10 + x 10 - x d x = ∫ 10 + x 10 - x 2 d x I = 10 ∫ 1 10 2 - x 2 d x + ∫ x 10 2 - x 2 d x Put 10 2 - x 2 = t ⇒ x d x = - d t 2 to the second integral, we get I = ∫ 1 10 2 - x 2 d x - 1 2 ∫ d t t I = ∫ 1 10 2 - x 2 d x - 1 2 ∫ t - 1 2 + c I = sin - 1 x 10 - 1 2 t 1 2 1 2 + c I = sin - 1 x 10 - 10 2 - x 2 + c

Evaluate the following ∫ 1 x 2 + 8 x + 12 · d x

∫ 1 x 2 + 8 x + 12 ⋅ d x = ∫ 1 ( x 2 + 8 x + 16 ) − 16 + 12 ⋅ d x = ∫ 1 ( x + 4 ) 2 − 2 2 ⋅ d x = 1 2 ( 2 ) log ⁡ | ( x + 4 ) − 2 ( x + 4 ) + 2 | + c = 1 4 log ⁡ | x + 2 x + 6 | + c .

Integrate the following functions w,r.t.x ∫ 1 3 + 2 sin 2 x + 4 cos 2 x · d x

I = ∫ 1 3 + 2 sin ⁡ 2 x + 4 cos ⁡ 2 x d x P u t tan ⁡ x = t ∴ x = tan − 1 ⁡ t ∴ dx = dt 1 + t 2 and sin ⁡ 2 x = 2 t 1 + t 2 , cos ⁡ 2 x = 1 − t 2 1 + t 2 ∴ I = ∫ 1 3 + 2 ( 2 t 1 + t 2 ) + 4 ( 1 − t 2 1 + t 2 ) ⋅ dt 1 + t 2 = ∫ 1 + t 2 3 ( 1 + t 2 ) + 4 t + 4 ( 1 − t 2 ) ⋅ dt 1 + t 2 = ∫ 1 7 + 4 t − t 2 dt = ∫ 1 7 − ( t 2 − 4 t + 4 ) + 4 dt = ∫ 1 ( 11 ) 2 − ( t − 2 ) 2 dt = 1 2 11 log ⁡ | 11 + t − 2 11 − t + 2 | + c = 1 2 11 log ⁡ | 11 + tan ⁡ x − 2 11 − tan ⁡ x + 2 | + c .

Integrate the following functions w,r.t.x ∫ 1 cos x - sin x · d x

Let I = ∫ 1 cos ⁡ x − sin ⁡ x ⋅ d x D i v i d i n g e a c h t e r m b y 1 2 + ( − 1 ) 2 = 2 , w e g e t I = 1 2 ∫ 1 cos ⁡ x ⋅ 1 2 − sin ⁡ x ⋅ 1 2 ⋅ d x = 1 2 ∫ 1 cos ⁡ x ⋅ cos ⁡ π 4 − sin ⁡ x ⋅ sin ⁡ π 4 ⋅ d x = 1 2 ∫ 1 cos ⁡ ( x + π 4 ) ⋅ d x = 1 2 ∫ sec ⁡ ( x + π 4 ) ⋅ d x = 1 2 log ⁡ | sec ⁡ ( x + π 4 ) + tan ⁡ ( x + π 4 ) | + c .

Integrate the following functions w,r.t.x ∫ 1 cos x - 3 sin x · d x

Let I = ∫ 1 cos ⁡ x − 3 sin ⁡ x ⋅ d x D i v i d i n g e a c h t e r m b y 1 2 + ( − 1 ) 2 = 2 , w e g e t I = 1 2 ∫ 1 cos ⁡ x ⋅ 1 2 − sin ⁡ x ⋅ 3 2 ⋅ d x = 1 2 ∫ 1 cos ⁡ x ⋅ cos ⁡ π 3 − sin ⁡ x ⋅ sin ⁡ π 3 ⋅ d x = 1 2 ∫ 1 cos ⁡ ( x + π 3 ) ⋅ d x = 1 2 ∫ sec ⁡ ( x + π 3 ) ⋅ d x = 1 2 log ⁡ | sec ⁡ ( x + π 3 ) + tan ⁡ ( x + π 3 ) | + c .

E M B I B E

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 3 - Indefinite Integration>EXERCISE 3.2 (B)

Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 3: EXERCISE 3.2 (B)

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 3: EXERCISE 3.2 (B)

Attempt the practice questions on Chapter 3: Indefinite Integration, Exercise 3: EXERCISE 3.2 (B) with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 2 Standard 12 solutions are prepared by Experienced Embibe Experts.