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

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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


\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}




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