If I=int_0^{frac{pi}{2}} frac{sin ^{frac{3}{2 | vedclass

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(A) Let $I_2 = \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we h

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

Vedclass · Answer

(A) Let $I_2 = \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I_2 = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin ^4(\frac{\pi}{2}-x)+\cos ^4(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2}-x) \cos x \sin x}{\cos ^4 x+\sin ^4 x} dx$. Adding the two expressions for $I_2$: $2I_2 = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. Divide numerator and denominator by $\cos^4 x$: $I_2 = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{	an x \sec^2 x}{	an^4 x + 1} dx$. Let $t = 	an^2 x$,then $dt = 2 	an x \sec^2 x dx$,so $	an x \sec^2 x dx = \frac{dt}{2}$. As $x 	o 0, t 	o 0$ and as $x 	o \frac{\pi}{2}, t 	o \infty$. $I_2 = \frac{\pi}{4} \int_0^{\infty} \frac{dt/2}{t^2+1} = \frac{\pi}{8} [	an^{-1} t]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$.

If $I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$,then the value of $\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$ is:

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