If \(I={\int }_{0}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx,\) the…

For \(I={\int }_{0}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx\)

Apply king's rule

\(I={\int }_{0}^{\frac{\pi }{2}}\frac{{\cos }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx\)

Adding both we get:

\(2\mathrm{I}={\int }_{0}^{\pi /2}\mathrm{dx}=\frac{\pi }{2}\Rightarrow \mathrm{I}=\frac{\pi }{4}\)

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

Apply king's rule

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

and adding both, we get

\(\text{ }{I}_{2}=\frac{\pi }{4}{\int }_{0}^{\pi /2}\frac{\tan x{\sec }^{2}xdx}{{\tan }^{4}x+1}\)

\(\text{ put }{\tan }^{2}\mathrm{x}=\mathrm{t}\Rightarrow \mathrm{tanx}{\sec }^{2}\mathrm{x}\mathrm{dx}=\frac{\mathrm{dt}}{2}\)

\({I}_{2}=\frac{\pi }{8}{\int }_{0}^{\infty }\frac{\mathrm{dt}}{{\mathrm{t}}^{2}+1}\)

\(=\frac{\pi }{8}{\left[{\tan }^{-1}\mathrm{t}\right]}_{0}^{\infty }\\ =\frac{\pi }{8}\cdot \frac{\pi }{2}=\frac{{\pi }^{2}}{16}\)