Mathematics - Definite Integration Question with Solution | TestHub

$f(t)={\int }_0^{\pi }\frac{2x}{1-{\cos }^{2}t{\sin }^{2}x}dx$ .....(1) $=2{\int }_0^{\pi }\frac{(\pi -x)dx}{1-{\cos }^2{\sin }^{2}x}$ .....(2) $\begin{array}{rl}& 2f(t)=2{\int }_0^{\pi }\frac{\pi }{1-{\cos }^2{\sin }^{2}x}dx\\ & f(t)={\int }_0^{\pi }\frac{\pi }{1-{\cos }^{2}t{\sin }^{2}x}dx\end{array}$ divide \& by ${\cos }^2\mathrm{x}$ $\begin{array}{rl}& f(t)=\pi {\int }_0^{\pi }\frac{{\sec }^{2}xdx}{{\sec }^{2}x-{\cos }^2{t}^{2}x}\\ & f(t)=2\pi {\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{{\sec }^{2}x-{\cos }^2{t}^2{\tan }^{2}x}\\ & \tan x=z\\ & {\sec }^{2}xdx=dz\\ & f(t)=2\pi {\int }_0^{\infty }\frac{dz}{1+{\sin }^{2}t\cdot z^2}\\ & =\frac{{\pi }^2}{\sin t}\end{array}$ Then ${\int }_0^{\pi /2}\frac{{\pi }^2}{f(t)}dt$ $\begin{array}{rl}& ={\int }_0^{\pi /2}\sin tdt\\ & =1\end{array}$