int_0^{pi /2} frac{x sin x cos x}{cos^4 x + s | vedclass

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(D) Let $I = \int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$ .....$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \,

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

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(D) Let $I = \int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$ .....$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^{\pi /2} \frac{(\frac{\pi}{2} - x) \cos x \sin x}{\sin^4 x + \cos^4 x} \, dx$ .....$(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^{\pi /2} \frac{\frac{\pi}{2} \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$Divide numerator and denominator by $\cos^4 x$:$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{	an x \sec^2 x}{1 + 	an^4 x} \, dx$Let $	an^2 x = t$,then $2 	an x \sec^2 x \, dx = dt$,so $	an x \sec^2 x \, dx = \frac{1}{2} dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t 	o \infty$.$I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt = \frac{\pi}{8} [	an^{-1} t]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$.

$\int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx = $

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