Question

${\int }_0^{\frac{\pi }{2}}\frac{x\mathrm{sinx}\mathrm{cosx}}{{\cos }^{4}x+{\sin }^{4}x}$ equals

• A. $\frac{{\pi }^2}{8}$
• B. $\frac{{\pi }^2}{16}$
• C. $\frac{{\pi }^2}{4}$
• D. $0$

Answer 1

Answer: (B)

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

$I={\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\pi /2}(0+\frac{\pi }{2}-x)\frac{\sin (\pi /2-x)\cos (\pi /2-x)}{{\sin }^4(\pi /2-x)+{\cos }^4(\pi /2-x)}dx$

$I={\int }_0^{\pi /2}\frac{\pi }{2}\left[\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\right]-{\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx-I$

$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$I=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{\sin 2x}{({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2}dx$

$=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{\sin 2x}{({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x}dx$

$I=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{\sin 2x}{1-2{\sin }^{2}x{\cos }^{2}x}=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{\sin 2xdx}{1-\frac{1}{2}\left({\sin }^{2}2x\right)}$

$=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\sin 2xdx}{2-{\sin }^{2}2x}$

$=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\sin 2xdx}{1+{\cos }^{2}2x}$

$\cos 2x=t$

$\Rightarrow 2\sin 2xdx=dt$

$I=\frac{-\pi }{4}{\int }_1^{-1}\frac{1}{2}\frac{dt}{(1+t^2)}$

$={\frac{-\pi }{8}{\tan }^{-1}t]}_1^{-1}$

$=\frac{-\pi }{8}[-\frac{\pi }{4}-\frac{\pi }{4}]=\frac{\pi }{8}\times \frac{\pi }{2}$

$I=\frac{{\pi }^2}{16}$.