f(t)=∫0π1−cos2tsin2x2xdx .....(1) =2∫0π1−cos2sin2x(π−x)dx .....(2) $\begin{aligned}
& 2 f(t)=2 \int_0^\pi \frac{\pi}{1-\cos ^2 \sin ^2 x} d x \
& f(t)=\int_0^\pi \frac{\pi}{1-\cos ^2 t \sin ^2 x} d x
\end{aligned}divide&by\cos ^2 \mathrm{x}\begin{aligned}
& f(t)=\pi \int_0^\pi \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 x} \
& f(t)=2 \pi \int_0^{\pi / 2} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 \tan ^2 x} \
& \tan x=z \
& \sec ^2 x d x=d z \
& f(t)=2 \pi \int_0^{\infty} \frac{d z}{1+\sin ^2 t \cdot z^2} \
& =\frac{\pi^2}{\sin t}
\end{aligned}Then\int_0^{\pi / 2} \frac{\pi^2}{f(t)} d t\begin{aligned}
& =\int_0^{\pi / 2} \sin t d t \
& =1
\end{aligned}$