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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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