The value of int_0^{pi /2} frac{sin x}{1 + co | vedclass

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(B) Let $I = \int_0^{\pi /2} \frac{\sin x}{1 + \cos^2 x} \, dx$.Substitute $\cos x = t$. Then,$-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.Change the

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$\pi /4$

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(B) Let $I = \int_0^{\pi /2} \frac{\sin x}{1 + \cos^2 x} \, dx$.Substitute $\cos x = t$. Then,$-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.Change the limits of integration:When $x = 0$,$t = \cos(0) = 1$.When $x = \pi /2$,$t = \cos(\pi /2) = 0$.Substituting these into the integral:$I = \int_1^0 \frac{-dt}{1 + t^2} = \int_0^1 \frac{dt}{1 + t^2}$.Using the standard integral $\int \frac{dt}{1 + t^2} = 	an^{-1} t$,we get:$I = [	an^{-1} t]_0^1 = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

The value of $\int_0^{\pi /2} \frac{\sin x}{1 + \cos^2 x} \, dx$ is

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