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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 17 Maths Textbook Solution.

Answer:$\frac{\pi }{4}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{2}}\cos ^{2}xdx$

Solution:$\int_{0}^{\frac{\pi }{2}}\cos ^{2}xdx$

$=\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\cos 2 x}{2}\right) d x \quad\left[\begin{array}{l} 2 \cos ^{2} \theta=1+\cos 2 \theta \\ \cos ^{2} \theta=\frac{1+\cos 2 \theta}{2} \end{array}\right]$

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}+\frac{\cos 2 x}{2}\right) d x=\int_{0}^{\frac{\pi}{2}} \frac{1}{2} d x+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x \end{aligned} \quad\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]

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

$=\frac{\pi }{4}+\frac{1}{4}\left [ \sin \pi -\sin 0 \right ]$

$=\frac{\pi }{4}$