Integrate the following functions w.r.t. x : sec 2 x · tan 2 x + tan x - 7

Let I = ∫ sec 2 x · tan 2 x + tan x - 7 d x Let t= tan x sec 2 x d x = d t I = ∫ t 2 + t - 7 d t = ∫ t 2 + t + 1 4 - 29 4 d t = ∫ t + 1 2 2 - 29 2 2 d t We know that ∫ x 2 - a 2 d x = x 2 x 2 - a 2 - a 2 2 ⋅ log x + x 2 - a 2 + c = t + 1 2 2 t + 1 2 2 - 29 2 2 - 29 2 2 2 ⋅ log t + 1 2 + t + 1 2 2 - 29 2 2 + c = t + 1 2 2 t 2 + t - 7 - 29 8 ⋅ log t + 1 2 + t 2 + t - 7 ∴ I = tan x + 1 2 2 tan 2 x + tan x - 7 - 29 8 ⋅ log tan x + 1 2 + tan 2 x + tan x - 7

Integrate the following functions w.r.t. x : x 2 + 2 x + 5

Let I = ∫ x 2 + 2 x + 5 d x = ∫ ( x 2 + 2 x + 1 ) + 4 d x = ∫ ( x + 1 ) 2 + 2 2 d x We know that ∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ⋅ log x + x 2 + a 2 + c = x + 1 2 ( x + 1 ) 2 + 2 2 + 2 2 2 ⋅ log ( x + 1 ) + ( x + 1 ) 2 + 2 2 + c ∴ I = x + 1 2 x 2 + 2 x + 5 + 2 log ( x + 1 ) + x 2 + 2 x + 5 + c

Integrate the following functions w.r.t. x : 2 x 2 + 3 x + 4

Let I = ∫ 2 x 2 + 3 x + 4 d x = ∫ 2 ( x 2 + 3 2 x + 2 ) d x = 2 ∫ ( x 2 + 3 2 x + 9 16 ) + 2 - 9 16 d x = 2 ∫ ( x + 3 4 ) 2 + 23 16 d x = 2 ∫ ( x + 3 4 ) 2 + 23 4 2 d x We know that ∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ⋅ log x + x 2 + a 2 + c = 2 x + 3 4 2 x + 3 4 2 + 23 4 2 + 23 4 2 2 ⋅ log x + 3 4 + x + 3 4 2 + 23 4 2 + c ∴ I = 2 x + 3 4 2 x 2 + 3 2 x + 2 + 23 32 log x + 3 4 + x 2 + 3 2 x + 2 + c

Integrate the following functions w.r.t. x : 1 + sin x 1 + cos x · e x

Let I = ∫ 1 + sin x 1 + cos x · e x d x We know that sin x = 2 sin x 2 cos x 2 1 + cos x = 2 cos 2 x 2 sin 2 x 2 + cos 2 x 2 = 1 I = ∫ e x sin 2 x 2 + cos 2 x 2 + 2 sin x 2 cos x 2 2 cos 2 x 2 d x = ∫ e x sin x 2 + cos x 2 2 2 cos 2 x 2 d x = 1 2 ∫ e x sin x 2 + cos x 2 cos x 2 2 d x = 1 2 ∫ e x tan x 2 + 1 2 d x = 1 2 ∫ e x tan 2 x 2 + 1 + 2 tan x 2 d x = 1 2 ∫ e x s e c 2 x 2 + 2 tan x 2 d x = ∫ e x 1 2 s e c 2 x 2 + tan x 2 d x L e t f ( x ) = tan x 2 t h e n f l ( x ) = 1 2 s e c 2 x 2 We know that ∫ e x f ( x ) + f l ( x ) d x = e x f ( x ) + c ∴ I = e x tan x 2 + c

Integrate the following functions w.r.t. x : e x x x log x 2 + 2 log x

Let I = ∫ e x x x log x 2 + 2 log x d x = ∫ e x log x 2 d x + 2 ∫ e x x log x d x We know that ∫ u . v d x = u ∫ v d x - ∫ d u d x . ∫ v d x d x + c Applying integration by parts in the first integral we get, I = log x 2 . e x - 2 ∫ log x x . e x d x + 2 ∫ log x x . e x d x + c where u = ( log x ) 2 a n d v = e x d u d x = 2 log x 1 x ∴ I = log x 2 . e x + c

Integrate the following functions w.r.t. x : e 5 x · 5 x · log x + 1 x

Let I = ∫ e 5 x · 5 x · log x + 1 x d x = ∫ e 5 x · 5 log x + 1 x d x P u t 5 x = t ⇒ 5 d x = d t ⇒ d x = d t 5 I = ∫ e t · 5 log t 5 + 5 t d t 5 = 1 5 ∫ e t · 5 log t 5 + 5 t d t Which is of the form ∫ e x · f ( x ) + f l ( x ) d x = e x f ( x ) + c Let f(x) = 5 log t 5 = 5 log t - 5 log 5 f l ( x ) = 5 t ∴ I = 1 5 e t . 5 log t 5 + c = e t . log t 5 + c Substituting t value we get I = e 5 x . log 5 x 5 + c ∴ I = e 5 x . log x + c

Integrate the following functions w.r. to x : log ( log x ) x

∫ ln ( ln ( x ) ) x d x Take: u = ln ( x ) ⇒ d u = 1 x d x = ∫ 1 . ln ( u ) d u Apply Integration By Parts: u = ln ( u ) , v ' = 1 = u ln ( u ) - ∫ 1 d u = u ln ( u ) - u + c = ln ( x ) ln ( ln ( x ) ) - ln ( x ) + c

E M B I B E

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 3 - Indefinite Integration>EXERCISE 3.3

Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 5: EXERCISE 3.3

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 5: EXERCISE 3.3

Attempt the practice questions on Chapter 3: Indefinite Integration, Exercise 5: EXERCISE 3.3 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 2 Standard 12 solutions are prepared by Experienced Embibe Experts.