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Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 12.

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\frac{\pi}{4}

Hint: We use indefinite integral formula then put limits to solve this integral.

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

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

Put \sin x=t

\cos x \; \; dx=dt

dx=\frac{dt}{\cos x}

When x=0  then t=0  and when x=\frac{\pi}{2}  then t=1

=\int_{0}^{1} \frac{1}{1+t^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]

=[\tan ^{-1}(t)]^{1}_0

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

=\left [\frac{\pi}{4}-0 \right ]

= \frac{\pi}{4}

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