Question

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

Answer 1

$I={\int }_0^{\frac{\pi }{2}}\frac{xsinxcosx}{si{n}^{4}x+co{s}^{4}x}$ --- (1)

$I={\int }_0^{\frac{\pi }{2}}\frac{\left(\right(\frac{\pi }{2}-x\left)cosxsinx\right)}{co{s}^4+si{n}^{4}x}dx$---(2)

Adding (1) and (2), we get

$2I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx$

Put $si{n}^{2}x=t$

$\Rightarrow 2sinxcosxdx=dt$

Therefore, $sinxcosxdx=\frac{dt}{2}$

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

$I=\frac{\pi }{8}{\int }_0^1\frac{dt}{2t^2-2t+1}\Rightarrow I=\frac{\pi }{16}\int \frac{dt}{t^2-t+\frac{1}{2}}$

$=\frac{\pi }{16}{\int }_0^1\frac{dt}{{(t-\frac{1}{2})}^2+\frac{1}{4}}={\begin{array}{c}\frac{\pi }{16}\frac{1}{\frac{1}{2}}ta{n}^{-1}\left(\frac{t-\frac{1}{2}}{\frac{1}{2}}\right)\end{array}|}_0^1=\frac{{\pi }^2}{16}$