int_0^pi frac{1}{1+sin x} , dx is equal to | vedclass

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(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} \, dx$. Using the identity $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$,we get: $I = \int_0^\pi \frac{1}{1

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$2$

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(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} \, dx$. Using the identity $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$,we get: $I = \int_0^\pi \frac{1}{1 + \frac{2 	an(x/2)}{1+	an^2(x/2)}} \, dx = \int_0^\pi \frac{1+	an^2(x/2)}{(1+	an(x/2))^2} \, dx$. Since $1+	an^2(x/2) = \sec^2(x/2)$,we have: $I = \int_0^\pi \frac{\sec^2(x/2)}{(1+	an(x/2))^2} \, dx$. Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) \, dx$,which implies $\sec^2(x/2) \, dx = 2 \, dt$. As $x 	o 0$,$t 	o 0$. As $x 	o \pi$,$t 	o \infty$. Thus,$I = \int_0^\infty \frac{2 \, dt}{(1+t)^2} = 2 \left[ -\frac{1}{1+t} ight]_0^\infty$. $I = 2 [0 - (-1)] = 2 	imes 1 = 2$.

$\int_0^\pi \frac{1}{1+\sin x} \, dx$ is equal to

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