#### explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 44 maths

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

Hint: Use substitution method to solve this integral.

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

Solution:

\begin{aligned} &\text { Let } I=\int \sec ^{4} x \tan x \; d x \\ &\text { Put } \tan x=t \Rightarrow \sec ^{2} x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{\sec ^{2} x} \text { then } \end{aligned}

$I=\int \sec ^{4} x \cdot t \cdot \frac{d t}{\sec ^{2} x}=\int \sec ^{2} x \cdot t\; d t$

$=\int\left\{1+\tan ^{2} x\right\} . t \; d t \quad\left[\begin{array}{l} \because \sec ^{2} x-\tan ^{2} x=1 \\ \Rightarrow \sec ^{2} x=1+\tan ^{2} x \end{array}\right]$

\begin{aligned} &=\int\left(1+t^{2}\right) . t\; d t \quad[\because \tan x=t] \\ &=\int\left(t+t^{2} t\right) d t=\int\left(t+t^{3}\right) d t \\ &=\int t \; d t+\int t^{3} d t \end{aligned}

$=\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

\begin{aligned} &=\frac{t^{2}}{2}+\frac{t^{4}}{4}+c \\ &=\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+c \quad[\because t=\tan x] \end{aligned}