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

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

Answers (1)

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}

 

 

 

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