Q. If $I=\int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}dx$, then $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}dx$ equals :
A) $\frac{\pi^2}{12}$
B) $\frac{\pi^2}4$
C) $\frac{\pi^2}{16}$
D) $\frac{\pi^2}8$
Explanation
[Solution by Dr. Abubacker] Let $J$ be the second integral. By $x\to\frac\pi2-x$ the weight $\sin x\cos x/(\sin^4x+\cos^4x)$ is unchanged, so $J=\frac\pi4\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$. With $u=\tan^2x$: $\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\frac12\int_0^\infty\frac{du}{1+u^2}=\frac\pi4$. So $J=\frac{\pi^2}{16}$. (The given $I$ is not needed.) Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.