#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 2 Maths Textbook Solution.

Answer: $\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C$

Hint :- Use substitution method to solve this integral.

Given :-   $\int \tan x \cdot \sec ^{4} x d x$

Sol : -  Let  $I=\int \tan x \sec ^{4} x d x$

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

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

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

Substitute $\tan x=t \rightarrow \sec ^{2} x d x=d t$ then

$\mathrm{I}=\int\left(\mathrm{t}+t^{3}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t} \text { ) }$

$=\int\left(\mathrm{t}+t^{3}\right) d t=\int \mathrm{t} d t+\int t^{3} d t$

$=\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+C$                                        $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{C}\right)$                        (i)

$I=\frac{t^{2}}{2}+\frac{t^{4}}{4}+C$

$I=\frac{\tan ^{2} x}{2}+\frac{\tan ^{4} x}{4}+\mathrm{C}$                        $\text { (if, } t=\tan x)$