Question

Evaluate :
${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{5}x}{{\sin }^{5}x+{\cos }^{5}x}\cdot dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{5}x}{{\sin }^{5}x+{\cos }^{5}x}\cdot dx$........(1)

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

$I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{5}x}{{\cos }^{5}x+{\sin }^{5}x}\cdot dx$........(2)

Since $\sin (\frac{\pi }{2}-\theta )=\cos \theta$

Now add (1) and (2)

$2I={\int }_0^{\frac{\pi }{2}}dx=\frac{\pi }{2}$

$\therefore I=\frac{\pi }{4}$