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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 13 maths textbook solution

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Answer: \sqrt{2}

Hint: You must know the integration rules of trigonometric function with its limits

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

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

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{2 \sin ^{2} x} \; d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{2} \sin x\; d x \end{aligned}

\begin{aligned} &=\sqrt{2}[-\cos x]_{0}^{\frac{\pi}{2}} \\\\ &=\sqrt{2}\left[-\cos \frac{\pi}{2}+\cos 0\right] \\\\ &=\sqrt{2} \end{aligned}


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