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  • Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 12

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 12

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Answer:  2(\sqrt{2-1})

Given:  \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{1+\cos x}} d x

Hint: Use substitution Method

Solution: \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{1+\cos x}} d x

Let  1+\cos x=t

-\sin x dx = dt   (diff w.r.t  x)

Now,

 \begin{aligned} &\int_{0}^{\frac{\pi}{2}}-\frac{1}{\sqrt{t}} d t \\ & \end{aligned}

=-\left(\frac{\sqrt{t}}{\frac{1}{2}}\right)_{0}^{\frac{\pi}{2}}=-2(\sqrt{1+\cos x})_{0}^{\frac{\pi}{2}}

\begin{aligned} &=-2(1-\sqrt{2}) \\ & \end{aligned}

=2(\sqrt{2}-1)

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