#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 53.

Hint: Use indefinite integral formula and the limits to solve this integral

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

Solution:

\begin{aligned} &\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2 \cos ^{2} \frac{x}{2}}}{\left(2 \sin ^{2} \frac{x}{2}\right)^{\frac{3}{2}}} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2} \cos \frac{x}{2}}{2^{\frac{3}{2}}\left(\sin ^{3} \frac{x}{2}\right)} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2} \cos \frac{x}{2}}{2 \sqrt{2}\left(\sin \frac{x}{2}\right)} \cdot \frac{1}{\left(\sin ^{2} \frac{x}{2}\right)} d x \\ &=\frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cot \frac{x}{2} \cos e c^{2} \frac{x}{2} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{\cos \theta}{\sin \theta}=\cot \theta, \frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] \end{aligned}

\begin{aligned} &\text { Put } \cot \frac{x}{2}=t \Rightarrow-\cos e c^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t\\ &\Rightarrow \operatorname{cosec}^{2} \frac{x}{2} d x=-2 d t\\ &\text { When } x=\frac{\pi}{3} \text { then } t=\sqrt{3} \& \text { when } x=\frac{\pi}{2} \text { then } t=1\\ &\therefore \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x=\frac{1}{2} \int_{\sqrt{3}}^{1} t \cdot(-2) d t \end{aligned}

\begin{aligned} &=-\int_{\sqrt{3}}^{1} t d t=-\left[\frac{t^{1+1}}{1+1}\right]_{\sqrt{3}}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=-\frac{1}{2}\left[t^{2}\right]_{\sqrt{3}}^{1} \\ &=-\frac{1}{2}\left[1^{2}-(\sqrt{3})^{2}\right] \\ &=-\frac{1}{2}[1-3] \\ &=\frac{-1}{2}[-2] \\ &=\frac{2}{2} \\ &=1 \end{aligned}