int_0^{frac{pi}{2}} frac{x tan x sec^2 x}{tan | vedclass

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(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{x 	an x \sec^2 x}{	an^4 x + 1} dx$  ...$(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we

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$\frac{\pi^2}{16}$

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(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{x 	an x \sec^2 x}{	an^4 x + 1} dx$  ...$(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get:$I = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) 	an(\frac{\pi}{2} - x) \sec^2(\frac{\pi}{2} - x)}{	an^4(\frac{\pi}{2} - x) + 1} dx$$I = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \cot x \csc^2 x}{\cot^4 x + 1} dx$Since $\cot x = \frac{1}{	an x}$ and $\csc^2 x = \frac{\sec^2 x}{	an^2 x}$, the expression becomes:$I = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \frac{1}{	an x} \cdot \frac{\sec^2 x}{	an^2 x}}{\frac{1}{	an^4 x} + 1} dx = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) 	an x \sec^2 x}{1 + 	an^4 x} dx$  ...(ii)Adding $(i)$ and (ii):$2I = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} 	an x \sec^2 x}{	an^4 x + 1} dx$$I = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{	an x \sec^2 x}{	an^4 x + 1} dx$Let $u = 	an^2 x$, then $du = 2 	an x \sec^2 x dx$, so $	an x \sec^2 x dx = \frac{du}{2}$.$I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{u^2 + 1} \cdot \frac{du}{2} = \frac{\pi}{8} [	an^{-1} u]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$

$\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx =$

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