Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 30 Maths Textbook Solution.

Answers (1)

Answer:

$\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c$

Given:

$\int \tan ^{5} x d x$

Hint:

To solve this statement we have to change tan into sec and then use formula such tan²x.

Solution:

$I=\int \tan ^{5} x d x$

$I=\int \tan ^{3} x \tan ^{2} x d x$

$=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x \quad\left[\because \tan ^{2} x=\sec ^{2} x-1\right]$

$=\int \tan ^{3} x \sec ^{2} x d x-\int \tan ^{3} x d x$

$=\int \tan ^{3} x \sec ^{2} x d x-\int \tan x \tan ^{2} x d x$

$I_{1}$ $I_{2}$

$\operatorname{let} \tan x=t, \sec ^{2} x d x=d t$

$=\int t^{3} d t-\int \tan x \tan ^{2} x d x$

$=\frac{t^{4}}{4}-\int \tan x\left(\sec ^{2} x-1\right) d x$

$=\frac{t^{4}}{4}-\int \tan x \sec ^{2} x d x+\int \tan x d x$

$=\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c$

$=\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c$

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount

anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question