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

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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