int_{0}^{pi} frac{x tan x}{sec x + tan x} dx | vedclass

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(D) Let $I = \int_{0}^{\pi} \frac{x 	an x}{\sec x + 	an x} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int

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

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(D) Let $I = \int_{0}^{\pi} \frac{x 	an x}{\sec x + 	an x} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi} \frac{(\pi - x) 	an(\pi - x)}{\sec(\pi - x) + 	an(\pi - x)} dx$. Since $	an(\pi - x) = -	an x$ and $\sec(\pi - x) = -\sec x$,we have: $I = \int_{0}^{\pi} \frac{(\pi - x)(-	an x)}{-\sec x - 	an x} dx = \int_{0}^{\pi} \frac{(\pi - x) 	an x}{\sec x + 	an x} dx$. $I = \pi \int_{0}^{\pi} \frac{	an x}{\sec x + 	an x} dx - I$. $2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx$. $2I = \pi \int_{0}^{\pi} \frac{\sin x(1 - \sin x)}{\cos^2 x} dx = \pi \int_{0}^{\pi} (\sec x 	an x - 	an^2 x) dx$. $2I = \pi \int_{0}^{\pi} (\sec x 	an x - (\sec^2 x - 1)) dx$. $2I = \pi [\sec x - 	an x + x]_{0}^{\pi}$. $2I = \pi [(\sec \pi - 	an \pi + \pi) - (\sec 0 - 	an 0 + 0)]$. $2I = \pi [(-1 - 0 + \pi) - (1 - 0 + 0)] = \pi (\pi - 2)$. $I = \frac{\pi (\pi - 2)}{2}$.

$\int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx =$

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