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Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 2

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Answer: \frac{\pi }{4}

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

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

Solution:  \int_{0}^{\frac{\pi}{2}} \cos ^{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 \frac{2 \pi}{2}-\sin 0\right] \\\\ &=\frac{\pi}{4} \end{aligned}


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