The value of int_0^{frac{pi}{2}} frac{sin x}{ | vedclass

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Question

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} dx$. Substitute $t = \cos x$,then $dt = -\sin x dx$,or $\sin x dx = -dt$. When $x = 0$,

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$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} dx$. Substitute $t = \cos x$,then $dt = -\sin x dx$,or $\sin x dx = -dt$. When $x = 0$,$t = \cos(0) = 1$. When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$. 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 $\int \frac{1}{1+t^2} dt = 	an^{-1} t$: $I = [	an^{-1} t]_0^1 = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

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

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