#### Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 20

Answer:   $\frac{3}{2}$

Given:  $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x$

Hint: Do rationalization and then apply substitution method

Solution:

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x$

\begin{aligned} &=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} \times \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} d x\\ & \end{aligned}

$=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1-\cos ^{2} x}}{(1-\cos x)^{2}} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sin x}{(1-\cos x)^{2}} d x$

\begin{aligned} &\text { Let } \\ & \end{aligned}

$1-\cos x=t \\$                          (differentiate w.r.t to x)

$\sin x d x=d t$

\begin{aligned} &=\int_{\frac{1}{2}}^{1} \frac{1}{t^{2}} d t=-\frac{1}{2}\left(t^{-1}\right)_{\frac{1}{2}}^{1}=-\frac{1}{4} \\ & \end{aligned}