# Get Answers to all your Questions

#### Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 14

The correct answer is $-e^{\frac{-x}{2}} \sec \left(\frac{x}{2}\right)+c$
Hint:

$I=\int e^{x}[f(x)+{f}'(x)]dx=e^{x}f(x)+c$

Given:

$\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{\frac{-x}{2}} d x$

Solution:

\begin{aligned} &I=\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{\frac{-x}{2}} d x \\ &\operatorname{Let}\left(\frac{-x}{2}\right)=t \quad x=-2 t \\ &\therefore d x=-2 d t \end{aligned}

$I=\int e^{t} \frac{\sqrt{(1-\sin (-2 t)}}{(1+\cos (-2 t)}(-2dt)$

$=\int e^{t} \frac{\sqrt{(} 1+\sin (2 t)}{(1+\cos (2 t)}(-2 d t)$

$=\int e^{t} \frac{(\cos t+\sin t)}{\cos ^{2} t}(-2 d t)$

$=(-1) \int e^{t}\left\{f(t)+f^{\prime}(t)\right\} d t$

Whose  $f(t)=\sec t$  and solution is given by $e^{t} f(t)+c$

\begin{aligned} &\therefore I=2e^{t} \operatorname{sect}(-1)+c \\ &\therefore I=-2e^{t} \sec t+c \end{aligned}

So, the correct answer is $-2e^{\frac{-x}{2}} \sec \left(\frac{x}{2}\right)+c$