int_0^{pi / 2} frac{16 x sin x cos x}{sin ^4 | vedclass

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(C) Let $I = \int_0^{\pi / 2} \frac{16 x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$  $(i)$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, w

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$\pi^2$

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(C) Let $I = \int_0^{\pi / 2} \frac{16 x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$  $(i)$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we have: $I = \int_0^{\pi / 2} \frac{16(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin^4(\frac{\pi}{2}-x)+\cos^4(\frac{\pi}{2}-x)} dx$ $I = \int_0^{\pi / 2} \frac{(8\pi - 16x) \cos x \sin x}{\cos^4 x + \sin^4 x} dx$  (ii) Adding $(i)$ and (ii): $2I = \int_0^{\pi / 2} \frac{8\pi \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $I = 4\pi \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$ Divide numerator and denominator by $\cos^4 x$: $I = 4\pi \int_0^{\pi / 2} \frac{	an x \sec^2 x}{	an^4 x + 1} dx$ Let $u = 	an^2 x$, then $du = 2 	an x \sec^2 x dx$: $I = 2\pi \int_0^{\infty} \frac{du}{u^2 + 1} = 2\pi [	an^{-1}(u)]_0^{\infty}$ $I = 2\pi [\frac{\pi}{2} - 0] = \pi^2$

$\int_0^{\pi / 2} \frac{16 x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$ is equal to

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