#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 119 Maths Textbook Solution.

Answer: $-e^{-x / 2} \sec \frac{x}{2}+C$

Hint : To solve this equation we have to use differentiation method

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

Solution:

$\int e^{-x}\left(f^{\prime}(x)-f(x) d x=e^{-x} f(x)+C\right.$

$\operatorname{Let} \frac{x}{2}=t$

$x=2 t$

$d x=2 d t$

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

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

$\frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t}=\frac{\sqrt{\sin ^{2} t+\cos ^{2} t-2 \sin t \cdot \cos t}}{2 \cos ^{2} t}$

$I=\frac{\sqrt{(\sin t-\cos t)^{2}}}{2 \cos ^{2} t}=\frac{\sin t-\cos t}{2 \cos ^{2} t}$

$=\frac{1}{2}\left(\frac{\sin t}{\cos ^{2} t}-\frac{\cos t}{\cos ^{2} t}\right)$

$=\frac{1}{2}\left(\frac{\tan t}{\cos t}-\frac{1}{\cos t}\right)$

$=\frac{1}{2}(\tan t \cdot \sec t-\sec t)$

$f(t)=\frac{1}{2} \sec t ; f^{\prime}(t)=\frac{1}{2} \tan t \sec t$

$\frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t}=f^{\prime}(t)-f(t)$

$I=-2 e^{-t} f(t)+C$

$I=-2 e^{-t} \cdot \frac{1}{2} \sec t+C$

$I=-e^{-x / 2} \sec \frac{x}{2}+C$