Evaluate the following integrals : ∫ x 3 - 2 x 2 + 5 e 3 x d x

We have to find out the given differential value. Let 3 x = t ⇒ x = t 3 3 d x = d t ⇒ ⇒ d x = d t 3 ∫ e t [ ( t 3 ) 3 - 2 ( t 3 ) 2 + 5 ] * d t 3 ∫ e t ( t 3 27 - 2 t 2 9 + 5 ) d t 3 1 3 * 27 ∫ e t ( t 3 + 3 t 2 - 9 t 2 - 18 t + 18 t + 18 + 117 ) d t 1 3 * 27 e t ( t 3 - 9 t 2 + 18 t + 117 ) Put value of t = 3 x 1 3 * 27 e 3 x [ ( 3 x ) 3 - 9 * ( 3 x ) 2 + 18 ( 3 x ) + 117 ] e 3 x 27 x 3 - 81 x 2 + 54 x + 117 3 * 27 e 3 x [ x 3 3 - x 2 + 2 3 x + 13 9 ] + C

Evaluate the following integrals : ∫ 1 + x 2 2 cos x d x

Integration by parts ⇒ I = 1 + x 2 2 ∫ cos x d x - ∫ d 1 + x 2 2 d x ∫ cos x d x ⇒ I = 1 + x 2 2 sin x - ∫ 2 1 + x 2 2 x sin x d x ⇒ I = 1 + x 2 2 sin x - ∫ 4 x sin x d x - ∫ 4 x 3 sin x d x ⇒ I = 1 + x 2 2 sin x - 4 - x cos x + sin x - 4 - x 3 cos x - ∫ 3 x 2 - cos x d x ⇒ I = 1 + x 2 2 sin x - 4 - x cos x + sin x - 4 - x 3 cos x + 3 x 2 sin x + 6 x cos x - 6 sin x + c ⇒ I = 1 + x 2 2 - 12 x 2 + 20 sin x + 4 x 3 - 20 x cos x + c

Evaluate the following integrals : ∫ x 2 + 2 x - 1 sin 3 x d x

Given question are- ∫ ( x 2 + 2 x - 1 ) sin 3 x d x Solve using integration by part method. ( x 2 + 2 x - 1 ) ∫ sin 3 x d x - ∫ d ( x 2 + 2 x - 1 ) d x ∫ sin 3 x ) d x d x ( x 2 + 2 x - 1 ) - cos 3 x 3 - ∫ ( 2 x + 2 ) - cos 3 x 3 d x ( x 2 + 2 x - 1 ) - cos 3 x 3 + 2 3 ∫ ( x + 1 ) cos 3 x d x ( x 2 + 2 x - 1 ) - cos 3 x 3 + 2 3 ( x + 1 ) ∫ cos 3 x d x - ∫ d ( x + 1 ) d x ∫ cos 3 x d x d x ( x 2 + 2 x - 1 ) - cos 3 x 3 + 2 3 ( x + 1 ) sin 3 x 3 - ∫ sin 3 x 3 d x ( x 2 + 2 x - 1 ) - cos 3 x 3 + 2 3 ( x + 1 ) sin 3 x 3 + cos 3 x 9 Ans.

Evaluate the following integrals : ∫ x 2 - 2 x + 3 ln x d x

Given integration are- ∫ x 2 - 2 x + 3 ln x d x Applying integration by part method- ∫ f ( x ) g ( x ) d x = f ( x ) ∫ g ( x ) d x - ∫ d f ( x ) d x ∫ g ( x ) d x d x Here, f ( x ) = ln x , g ( x ) = ( x 2 - 2 x + 3 ) Now, ln x ∫ ( x 2 - 2 x + 3 ) d x - ∫ d ln x d x ∫ ( x 2 - 2 x + 3 ) d x d x ln x x 3 3 - x 2 + 3 x - ∫ 1 x x 3 3 - x 2 + 3 x d x ln x x 3 3 - x 2 + 3 x - ∫ x 2 3 - x + 3 d x ln x x 3 3 - x 2 + 3 x - x 3 9 - x 2 2 + 3 x + c Ans.

Evaluate the following integrals : ∫ x 3 arc tan x d x

Given integration are- ∫ x 3 arc tan x d x ∫ x 3 tan - 1 x d x Apply integration by part method. tan - 1 x ∫ x 3 d x - ∫ d tan - 1 x d x ∫ x 3 d x d x tan - 1 x x 4 4 - ∫ 1 1 + x 2 x 4 4 d x tan - 1 x x 4 4 - 1 4 ∫ x 4 1 + x 2 d x tan - 1 x x 4 4 - 1 4 ∫ x 2 + 1 x 2 - 1 + 1 1 + x 2 d x tan - 1 x x 4 4 - 1 4 ∫ x 2 - 1 + 1 1 + x 2 d x tan - 1 x x 4 4 - 1 4 ∫ x 2 - 1 d x + ∫ 1 1 + x 2 d x tan - 1 x x 4 4 - 1 4 x 3 3 - x + tan - 1 x + c Ans.

Evaluate the following integrals : ∫ x 2 arc cos x d x

Given integration are- ∫ x 2 arc cos x d x ∫ x 2 cos - 1 x d x Solve using integration by part method, = cos - 1 x ∫ x 2 d x - ∫ d cos - 1 x d x ∫ x 2 d x d x = cos - 1 x x 3 3 - ∫ - 1 1 - x 2 x 3 3 d x = x 3 3 cos - 1 x + 1 3 ∫ x 3 1 - x 2 d x Let y = 1 - x 2 ⇒ d y = - 2 x d x Now, x 3 3 cos - 1 x + 1 3 ∫ 1 - y y - d y 2 x 3 3 cos - 1 x + 1 6 ∫ y - 1 y d y x 3 3 cos - 1 x + 1 6 ∫ y 1 2 d y - 1 6 ∫ y - 1 2 x 3 3 cos - 1 x + 1 6 y 3 2 3 2 - 1 6 y 1 2 1 2 + c x 3 3 cos - 1 x + 1 9 y 3 2 - 1 3 y 1 2 + c x 3 3 cos - 1 x + 1 9 1 - x 2 3 2 - 1 3 1 - x 2 1 2 + c Ans.

E M B I B E

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 4 - Indefinite Integrals, Basic Methods of Integration>Integration by Parts

I A Maron Solutions for Chapter: Indefinite Integrals, Basic Methods of Integration, Exercise 3: Integration by Parts

Author:I A Maron

I A Maron Mathematics Solutions for Exercise - I A Maron Solutions for Chapter: Indefinite Integrals, Basic Methods of Integration, Exercise 3: Integration by Parts

Attempt the practice questions on Chapter 4: Indefinite Integrals, Basic Methods of Integration, Exercise 3: Integration by Parts with hints and solutions to strengthen your understanding. PROBLEMS IN CALCULUS OF ONE VARIABLE solutions are prepared by Experienced Embibe Experts.