Evaluate the integral int_{0}^{frac{pi}{2}} f | vedclass

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Question

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.Substitute $\cos x = t$,which implies $-\sin x \,d x = d t$ or $\sin x \,d x

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.Substitute $\cos x = t$,which implies $-\sin x \,d x = d t$ or $\sin x \,d x = -d t$.Change the limits of integration:When $x = 0$,$t = \cos(0) = 1$.When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$.Now,substitute these into the integral:$I = \int_{1}^{0} \frac{-d t}{1+t^{2}} = \int_{0}^{1} \frac{d t}{1+t^{2}}$.Using the standard integral formula $\int \frac{1}{1+t^2} \,dt = 	an^{-1} t + C$:$I = [	an^{-1} t]_{0}^{1}$.Evaluate at the limits:$I = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.

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