Question

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

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

Answer 1

The correct option is D $\frac{{\pi }^2}{16}$

$I={\int }_0^{\pi /2}\frac{xsinxcosx}{co{s}^{4}x+si{n}^{4}x}dx\dots \left(i\right)={\int }_0^{\pi /2}\frac{(\frac{\pi }{2}-x)cosxsinx}{si{n}^{4}x+co{s}^{4}x}dx\dots \left(ii\right)$ By adding (i) and (ii), we get$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{cosxsinx}{co{s}^{4}x+si{n}^{4}x}dx\Rightarrow I=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{tanxse{c}^{2}x}{1+ta{n}^{4}x}dx$Now, put $ta{n}^{2}x=t$, we get
$I=\frac{\pi }{4}{\int }_0^{\infty }\frac{dt}{2(1+t^2)}=\frac{\pi }{8}{\left[ta{n}^{-1}t\right]}_0^{\infty }=\frac{{\pi }^2}{16}$