If f(t) = int_0^pi frac{2x , dx}{1 - cos^2 t | vedclass

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(C) Given $f(t) = \int_0^\pi \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$.Using the property $\int_0^a g(x) \, dx = \int_0^a g(a-x) \, dx$,we have:$f(t) =

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(C) Given $f(t) = \int_0^\pi \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$.Using the property $\int_0^a g(x) \, dx = \int_0^a g(a-x) \, dx$,we have:$f(t) = \int_0^\pi \frac{2(\pi - x) \, dx}{1 - \cos^2 t \sin^2 x}$.Adding the two expressions for $f(t)$:$2f(t) = \int_0^\pi \frac{2x + 2\pi - 2x}{1 - \cos^2 t \sin^2 x} \, dx = \int_0^\pi \frac{2\pi \, dx}{1 - \cos^2 t \sin^2 x}$.Thus,$f(t) = \pi \int_0^\pi \frac{dx}{1 - \cos^2 t \sin^2 x}$.Since the integrand is symmetric about $x = \frac{\pi}{2}$,$f(t) = 2\pi \int_0^{\frac{\pi}{2}} \frac{dx}{1 - \cos^2 t \sin^2 x}$.Dividing numerator and denominator by $\cos^2 x$:$f(t) = 2\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{\sec^2 x - \cos^2 t 	an^2 x} = 2\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + 	an^2 x - \cos^2 t 	an^2 x} = 2\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + \sin^2 t 	an^2 x}$.Let $	an x = z$,then $\sec^2 x \, dx = dz$. As $x 	o 0, z 	o 0$ and as $x 	o \frac{\pi}{2}, z 	o \infty$:$f(t) = 2\pi \int_0^{\infty} \frac{dz}{1 + (\sin t \cdot z)^2} = 2\pi \left[ \frac{1}{\sin t} 	an^{-1}(\sin t \cdot z) ight]_0^{\infty} = 2\pi \cdot \frac{1}{\sin t} \cdot \frac{\pi}{2} = \frac{\pi^2}{\sin t}$.Now,calculate the integral $\int_0^{\frac{\pi}{2}} \frac{\pi^2}{f(t)} \, dt = \int_0^{\frac{\pi}{2}} \frac{\pi^2}{\pi^2 / \sin t} \, dt = \int_0^{\frac{\pi}{2}} \sin t \, dt$.$= [-\cos t]_0^{\frac{\pi}{2}} = -(0 - 1) = 1$.

If $f(t) = \int_0^\pi \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$,where $0 < t < \pi$,then the value of $\int_0^{\frac{\pi}{2}} \frac{\pi^2 \, dt}{f(t)}$ equals..........

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