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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

Answers (1)

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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