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

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

Answers (1)

Answer: 1  

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}

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