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  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 29 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 29 maths

Answers (1)

Answer:

2(\sqrt{2}-1)

Hint:

First integrate and then use limits

Given:

\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} \; d x

Solution:

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \sqrt{|\sin x-\cos x|^{2}} d x \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x \\\\ &=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x)dx-\int_{0}^{\frac{\pi}{4}}(\sin x-\cos x)dx \end{aligned}

\begin{aligned} &I=[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-|-\cos x-\sin x|_{0}^{\frac{\pi}{4}} \\\\ &I=-0-1+\sqrt{2}-(-\sqrt{2}+1)=2 \sqrt{2}-2 \\\\ &I=2(\sqrt{2}-1) \end{aligned}

 

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