Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& I={\int }_0^{\frac{\pi }{2}}\frac{96x^2{\cos }^{2}x}{1+e^x}dx\\ & 2I=2{\int }_0^{\frac{\pi }{2}}96x^2{\cos }^{2}xdx\\ & I=96{\int }_0^{\frac{\pi }{2}}{x}^2{\cos }^{2}xdx\\ & =48{\int }_0^{\frac{\pi }{2}}{x}^2(1+\cos 2x)dx\\ & =2{\pi }^2+48(0-0)-48{\int }_0^{\frac{\pi }{2}}x\sin 2xdx\\ & =2{\pi }^2-12\pi +[0-0]=\pi \left(2{\pi }^2-12\right)\\ & =\pi \left(\alpha {\pi }^2+\beta \right)\\ & \Rightarrow (\alpha +\beta {)}^2=100\end{array}$