Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& I={\int }_0^{\frac{\pi }{2}}\frac{(\sin x{)}^{\frac{3}{2}}dx}{(\sin x{)}^{\frac{3}{2}}x+(\cos x{)}^{\frac{3}{2}}}={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}\left(\frac{\pi }{2}-x\right)dx}{{\sin }^{\frac{3}{2}}\left(\frac{\pi }{2}-x\right)+{\cos }^{\frac{3}{2}}\left(\frac{\pi }{2}-x\right)}\\ & \Rightarrow \text{ Adding }2l={\int }_0^{\frac{\pi }{2}}\frac{(\sin x{)}^{\frac{3}{2}}+(\cos x{)}^{\frac{3}{2}}}{(\sin x{)}^{\frac{3}{2}}+(\cos x{)}^{\frac{3}{2}}}dx=\frac{\pi }{2}\\ & I_0={\int }_0^{\frac{\pi }{2}}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx={\int }_0^{\frac{\pi }{2}}\frac{\left(\frac{\pi }{2}-x\right)\sin x\cos x}{(\sin x{)}^4+(\cos x{)}^4}dx\end{array}$ Adding, $2I_0={\int }_0^{\frac{\pi }{2}}\frac{\frac{\pi }{2}(\sin x)\cos x}{\left({\sin }^{4}x+{\cos }^{4}x\right)}dx$ $\Rightarrow I_0=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{\tan x\left({\sec }^{2}x\right)dx}{1+{\tan }^{4}x}$ put ${\tan }^{2}x=t\Rightarrow 2\tan x{\sec }^{2}xdx=dt$ $\begin{array}{rl}& \Rightarrow I_0=\frac{\pi }{4}{\int }_0^{\infty }\frac{\frac{dt}{2}}{\left(1+t^2\right)}={\frac{\pi }{8}\left({\tan }^{-1}t\right)|}_0^{\infty }=\frac{\pi }{8}\left(\frac{\pi }{2}-0\right)\\ & \Rightarrow I_0=\frac{{\pi }^2}{16}\end{array}$