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

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(C) Let $I = \int_0^\pi \frac{x \, dx}{1 + \sin x}$ ....$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:$I = \int_0^\p

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$\pi$

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(C) Let $I = \int_0^\pi \frac{x \, dx}{1 + \sin x}$ ....$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:$I = \int_0^\pi \frac{(\pi - x) \, dx}{1 + \sin(\pi - x)}$$I = \int_0^\pi \frac{(\pi - x) \, dx}{1 + \sin x}$ ....$(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi \frac{x + \pi - x}{1 + \sin x} \, dx = \pi \int_0^\pi \frac{1}{1 + \sin x} \, dx$Multiply numerator and denominator by $(1 - \sin x)$:$2I = \pi \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} \, dx = \pi \int_0^\pi \frac{1 - \sin x}{\cos^2 x} \, dx$$2I = \pi \int_0^\pi (\sec^2 x - \sec x 	an x) \, dx$Integrating the terms:$2I = \pi [	an x - \sec x]_0^\pi$$2I = \pi [(	an \pi - \sec \pi) - (	an 0 - \sec 0)]$$2I = \pi [(0 - (-1)) - (0 - 1)]$$2I = \pi [1 + 1] = 2\pi$$I = \pi$

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

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