Correct answer: 1
\(f(t)=\int\limits_{0}^{\pi} \frac{2 x}{1-\cos ^{2} \sin ^{2} x} d x\quad ....(1)\)
\(=2 \int\limits_{0}^{\pi} \frac{(\pi-x) d x}{1-\cos ^{2} t \sin ^{2} x} \quad ....(2)\)
\(2 f(t)=2 \int\limits_{0}^{\pi} \frac{\pi}{1-\cos ^{2} t \sin ^{2} x} d x\)
\(f(t)=\int\limits_{0}^{\pi} \frac{\pi}{1-\cos ^{2}t \sin ^{2} x} d x\)
divide & by \(\cos ^{2} \mathrm{x}\)
\(f(t)=\pi \int\limits_{0}^{\pi} \frac{\sec ^{2} x d x}{\sec ^{2} x-\cos ^{2} t \tan ^{2} x}\)
\(f(t)=2 \pi \int\limits_{0}^{\pi / 2} \frac{\sec ^{2} x d x}{\sec ^{2} x-\cos ^{2} t \tan ^{2} x}\)
\(\tan \mathrm{x}=\mathrm{z}\)
\(\sec ^{2} x d x=d z\)
\(f(t)=2 \pi \int\limits_{0}^{\infty} \frac{d z}{1+\sin ^{2} t \cdot z^{2}}\)
\(=\frac{\pi^{2}}{\sin t}\)
Then \(\int\limits_{0}^{\pi / 2} \frac{\pi^{2}}{f(t)} d t\)
\(=\int\limits_{0}^{\pi / 2} \sin t d t\)
\(=1\)
