Evaluate : ∫ d x ( x + 3 - x + 2 )

Given, ∫ d x ( x + 3 - x + 2 ) Multiply and divide by ( x + 3 ) + ( x + 2 ) = ∫ d x ( x + 3 - x + 2 ) × ( x + 3 + x + 2 ) ( x + 3 + x + 2 ) d x = ∫ ( ( x + 3 ) + ( x + 2 ) ) x + 3 - x - 2 d x = ∫ ( x + 3 ) d x + ∫ ( x + 2 ) d x = ( x + 3 ) 3 2 3 2 + ( x + 2 ) 3 2 3 2 + c ∴ ∫ d x ( x + 3 - x + 2 ) = 2 3 ( x + 3 ) 3 2 + 2 3 ( x + 2 ) 3 2 + c .

Evaluate : ∫ 1 + cos x 1 - cos x d x

Given, ∫ 1 + cos x 1 - cos x d x Multiply and divide by ( 1 + cos x ) = ∫ ( 1 + cos x ) 2 1 - ( cos x ) 2 d x = ∫ 1 + ( cos x ) 2 + 2 cos x ( sin x ) 2 d x ∵ sin 2 x + cos 2 x = 1 = ∫ ( cosec x ) 2 + ( cot x ) 2 + 2 cot x cosec x d x = ∫ cosec 2 x + cosec 2 x - 1 + 2 cot x cosec x d x ∵ cosec 2 x - cot 2 x = 1 = 2 ∫ cosec 2 x d x - 2 ∫ cot x cosec x d x - ∫ 1 d x ∴ ∫ 1 + cos x 1 - cos x d x = - 2 cot x - 2 cosec x - x + c

Evaluate : ∫ ( 1 + tan x ) ( 1 - tan x ) d x

Let I = ∫ ( 1 + tan x ) ( 1 - tan x ) d x = ∫ 1 + sin x cos x 1 - sin x cos x d x = ∫ cos x + sin x cos x - sin x d x Let cos x - sin x = t ⇒ - ( sin x + cos x ) d x = d t ⇒ ( sin x + cos x ) d x = - d t So, I = - ∫ d t t = - ln | t | + c ∴ I = - ln | cos x - sin x | + c .

Evaluate : ∫ cos ( x + a ) sin ( x + b ) d x

Given, ∫ cos ( x + a ) sin ( x + b ) d x = ∫ cos ( x + b + a - b ) sin ( x + b ) d x = ∫ cos x + b + a - b sin ( x + b ) d x = ∫ cos ( x + b ) cos ( a - b ) - sin ( x + b ) sin ( a - b ) sin ( x + b ) d x ∵ cos A + B = cos A cos B - sin A sin B = cos ( a - b ) ∫ cos( x + b ) sin ( x + b ) d x - sin ( a - b ) ∫ d x = cos ( a - b ) · ln | sin ( x + b ) | - x sin ( a - b ) + c ∵ ∫ cot x d x = log sin x + c ∴ ∫ cos ( x + a ) sin ( x + b ) d x = cos ( a - b ) · ln | sin ( x + b ) | - x sin ( a - b ) + c .

Evaluate : ∫ sin ( x - α ) sin ( x + α ) d x

Given, ∫ sin ( x - α ) sin ( x + α ) d x = ∫ sin ( x + α - α - α ) sin ( x + α ) d x = ∫ sin ( x + α ) cos ( 2 α ) - sin ( 2 α ) cos ( x + α ) sin ( x + α ) d x ∵ sin A - B = sin A cos B - cos A sin B = ∫ cos 2 α d x - sin 2 α ∫ cos ( x + α ) sin x + α d x ∵ ∫ cot x d x = log sin x + c ∵ ∫ sin ( x - α ) sin ( x + α ) d x = x cos 2 α - sin 2 α · ln | sin ( x + α ) | + c .

Evaluate : ∫ ( 1 - x ) x d x

Given, ∫ ( 1 - x ) x d x = ∫ x 1 2 - x 3 2 d x = ∫ x 1 2 d x - ∫ x 3 2 d x = x 3 2 3 2 - x 5 2 5 2 + c ∴ ∫ ( 1 - x ) x d x = 2 3 x 3 2 - 2 5 x 5 2 + c .

Evaluate : ∫ sec 2 x cosec 2 x d x

Given, ∫ sec 2 x cosec 2 x d x = ∫ sin 2 x cos 2 x d x = ∫ tan 2 x d x = ∫ sec 2 x - 1 d x ∵ sec 2 x - tan 2 x = 1 = ∫ sec 2 x d x - ∫ d x ∴ ∫ sec 2 x cosec 2 x d x = tan x - x + c .

Evaluate : ∫ 2 - 3 sin x cos 2 x d x

Given, ∫ 2 - 3 sin x cos 2 x d x = ∫ 2 ( cos x ) 2 d x - ∫ 3 sin x ( cos x ) 2 d x = ∫ 2 sec 2 x d x - ∫ 3 tan x sec x d x = 2 ∫ ( sec x ) 2 d x - 3 ∫ tan x sec x d x ∴ ∫ 2 - 3 sin x cos 2 x d x = 2 tan x - 3 sec x + c .

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SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Integral Calculus>Chapter 12 - Indefinite Integral>EXERCISE

R S Aggarwal Solutions for Exercise 1: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE

Attempt the free practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.