Question

If $I={\int }_0^{\pi /2}\frac{{\sin }^{3}x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

then $I$ is equal to?

• A. $\pi /8$
• B. $\pi /4$
• C. $\pi /2$
• D. $\pi$

Answer 1

The correct option is A $\pi /8$

Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$ we get
$I={\int }_0^{\pi /2}\frac{{\cos }^{3}x\sin x}{{\cos }^{4}x+{\sin }^{4}x}dx\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right),$ we get
$2I={\int }_0^{\pi /2}\frac{\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$
Using $co{s}^{4}x+{\sin }^{4}x=(co{s}^{2}x{)}^2+(si{n}^{2}x{)}^2$
$=\left(1/4\right)(1-cos2x{)}^2+(1/4\left)\right(1+\cos 2x{)}^2$
$=\left(1/2\right)(1+{\cos }^{2}2x)$
$\therefore 2I={\int }_0^{\pi /2}\frac{2\sin xcosx}{1+{\cos }^{2}2x}dx$
Substitute $\cos 2x=t,$ so that
$2I=\frac{1}{2}{\int }_1^{-1}\frac{dt}{1+t^2}=-{\int }_0^1\frac{dt}{1+t^2}=\frac{\pi }{4}$

$\Rightarrow I=\pi /8$