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Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 13.

Answers (1)

Answer: 2(\sqrt{2}-1)

Hint: We use indefinite integral formula then put limits to solve this integral.

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

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

Put 1+cos\theta =t^2

-\sin \theta \; \; d\theta=2t\; dt

When \theta=0  then t=2  and

When  \theta=\frac{\pi}{2}  then t=1

\begin{aligned} I &=\int_{\sqrt{2}}^{1} \frac{-2 t d t}{\sqrt{t^{2}}} d t \\ &=-\int_{\sqrt{2}}^{1} \frac{2 t}{t} d t \\ &=-2 \int_{\sqrt{2}}^{1} 1 d t \\ &=-2[t]_{\sqrt{2}}^{1} \\ &=-2[1-\sqrt{2}] \\ &=2(\sqrt{2}-1) \end{aligned}

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