If $f(t)=\int_0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0 < \mathrm{t} < \pi$, then…
If $f(t)=\int_0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0 < \mathrm{t} < \pi$, then the value of $\int_0^{\frac{\pi}{2}} \frac{\pi^2 \mathrm{dt}}{f(\mathrm{t})}$ equals_________
Solution
$f(t)=\int_0^\pi \frac{2 x}{1-\cos ^2 t \sin ^2 x} d x$ .....(1)
$=2 \int_0^\pi \frac{(\pi-x) d x}{1-\cos ^2 \sin ^2 x}$ .....(2)
$\begin{aligned}
& 2 f(t)=2 \int_0^\pi \frac{\pi}{1-\cos ^2 \sin ^2 x} d x \\
& f(t)=\int_0^\pi \frac{\pi}{1-\cos ^2 t \sin ^2 x} d x
\end{aligned}$
divide \& by $\cos ^2 \mathrm{x}$
$\begin{aligned}
& f(t)=\pi \int_0^\pi \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 x} \\
& f(t)=2 \pi \int_0^{\pi / 2} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 \tan ^2 x} \\
& \tan x=z \\
& \sec ^2 x d x=d z \\
& f(t)=2 \pi \int_0^{\infty} \frac{d z}{1+\sin ^2 t \cdot z^2} \\
& =\frac{\pi^2}{\sin t}
\end{aligned}$
Then $\int_0^{\pi / 2} \frac{\pi^2}{f(t)} d t$
$\begin{aligned}
& =\int_0^{\pi / 2} \sin t d t \\
& =1
\end{aligned}$