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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 1 maths textbook solution

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Answer: \frac{\pi }{4}

Hint: You must know the integration rule of trigonometric functions with its limits.

Given: \int_{0}^{\frac{\pi}{2}} \sin ^{2} x\; d x


\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x \\\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 . d x-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x \; d x \end{aligned}

\begin{aligned} &=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{4}\left[\sin \left(2 \frac{\pi}{2}\right)-\sin 0\right] \end{aligned}

\begin{aligned} &=\frac{\pi}{4}-\frac{1}{4}[0] \\\\ &=\frac{\pi}{4} \end{aligned}


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