#### explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 16 maths

Answer: $2 \sqrt{\tan x}+c$

Hint: Use substitution method to solve this integral.

Given: $\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$

Solution:

\begin{aligned} &I=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x \\ &=\int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cdot \cos x \cdot \cos x} d x \end{aligned}

$=\int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cdot \cos ^{2} x} d x=\int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x}} \cdot \frac{1}{\cos ^{2} x} d x$

\begin{aligned} &=\int \frac{\sqrt{\tan x}}{\tan x} \cdot \sec ^{2} x\; d x=\int(\tan x)^{\frac{1}{2}-1} \cdot \sec ^{2} x\; d x \\ &=\int(\tan x)^{\frac{-1}{2}} \sec ^{2} x \; d x \end{aligned}

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

$I=\int t^{\frac{-1}{2}} d t=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

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