Evaluate the definite integral: int_0^pi frac | vedclass

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(A) Let $I = \int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get: $I = \int_0^\pi \frac{(\p

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$\frac{\pi(\pi-2)}{2}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get: $I = \int_0^\pi \frac{(\pi-x) \cos^2(\pi-x)}{1+\sin(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \cos^2 x}{1+\sin x} dx$. Adding the two expressions for $I$: $2I = \int_0^\pi \frac{\pi \cos^2 x}{1+\sin x} dx = \pi \int_0^\pi \frac{1-\sin^2 x}{1+\sin x} dx$. Since $1-\sin^2 x = (1-\sin x)(1+\sin x)$, we have: $2I = \pi \int_0^\pi (1-\sin x) dx$. $2I = \pi [x + \cos x]_0^\pi$. $2I = \pi [(\pi + \cos \pi) - (0 + \cos 0)] = \pi [(\pi - 1) - (1)] = \pi(\pi - 2)$. Therefore, $I = \frac{\pi(\pi-2)}{2}$.

Evaluate the definite integral: $\int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$

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