int_0^{pi / 2} frac{sin ^3 x cos x , dx}{sin | vedclass

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(D) Let $I = \int_0^{\pi / 2} \frac{\sin ^3 x \cos x}{\sin ^4 x+\cos ^4 x} \, dx \quad \dots (i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(

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$\frac{\pi}{8}$

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(D) Let $I = \int_0^{\pi / 2} \frac{\sin ^3 x \cos x}{\sin ^4 x+\cos ^4 x} \, dx \quad \dots (i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:$I = \int_0^{\pi / 2} \frac{\sin ^3(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin ^4(\frac{\pi}{2}-x)+\cos ^4(\frac{\pi}{2}-x)} \, dx$$I = \int_0^{\pi / 2} \frac{\cos ^3 x \sin x}{\cos ^4 x+\sin ^4 x} \, dx \quad \dots (ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^{\pi / 2} \frac{\sin x \cos x (\sin ^2 x + \cos ^2 x)}{\sin ^4 x+\cos ^4 x} \, dx$$2I = \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} \, dx$Divide numerator and denominator by $\cos^4 x$:$2I = \int_0^{\pi / 2} \frac{	an x \sec^2 x}{	an ^4 x+1} \, dx$Let $	an^2 x = t$,then $2 	an x \sec^2 x \, dx = dt$,so $	an x \sec^2 x \, dx = \frac{dt}{2}$.When $x=0, t=0$ and when $x=\frac{\pi}{2}, t 	o \infty$.$2I = \int_0^{\infty} \frac{1}{t^2+1} \cdot \frac{dt}{2} = \frac{1}{2} [	an^{-1} t]_0^{\infty} = \frac{1}{2} (\frac{\pi}{2} - 0) = \frac{\pi}{4}$.$2I = \frac{\pi}{4} \implies I = \frac{\pi}{8}$.

$\int_0^{\pi / 2} \frac{\sin ^3 x \cos x \, dx}{\sin ^4 x+\cos ^4 x} = ?$

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