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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 13

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Answer: \frac{\pi}{4}

Given:  \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin ^{2} x} d x

Hint: You must know about the integration of  \frac{1}{1+x^{2}}

Solution:  \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin ^{2} x} d x

Let \sin x=t

\cos xdx=dt    (Differentiate w.r.t to x)

Now  \int_{0}^{\frac{\pi}{2}} \frac{1}{1+t^{2}} d t

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

=\tan ^{-1} 1-\tan ^{-1} 0 \\

=\frac{\pi}{4}-0\left(\because \tan \frac{\pi}{4}=1\right)

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