#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 5 Maths Textbook Solution.

Answer:$\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec \times| C$

Hint:- Use substitution method to solve this integral.

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

Solution: Let,$\mathrm{I}=\int \tan ^{5} x d x$

Re-Write,$\mathrm{I}=\int \tan ^{3} x \cdot \tan ^{2} x \cdot d x$

$I=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$            $\text { (if, } \left.\tan ^{2} x=\sec ^{2} x-1\right)$

$\mathrm{I}=\int\left(\tan ^{3} x \cdot \sec ^{2} x-\tan ^{3}\right) d x$

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

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

$\inline \left.=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x-1\right) \cdot \tan x d x \quad \text { (if, } \sec ^{2} x-\tan ^{2} x=1\right)$

$\inline =\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x \tan x-\tan x\right) d x$

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

Put    $tan x = t \Rightarrow \sec^{2}x d x = d t , then$

$\mathrm{I}=\int \mathrm{t}^{3} d t-\int t d t+\int \tan x d x$

$=\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sec \times| C$        $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \tan x d x=\log |\sec x|+c\right)$

$=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x| C$

$\left.=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x| C\: \: \: \: \: \: \quad \text { (if, } t=\tan x\right)$

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