Animated Solution for Mathematics - Definite Integration: If $f(t)={\int }_0^{\pi }\frac{2xdx}{1-{\cos }^{2}t{\sin }^{2}x},0<t<\pi$, then the value of ${\int }_0^{\pi /2}\frac{{\pi }^{2}dt}{f(t)}$ equals

Welcome, fellow traveler of the mathematical realms. Today, we are going to dismantle a problem that, at first glance, looks like a formidable fortress.

We are presented with the function:

That isolated $x$ in the numerator is a classic 'troublemaker.' It sits there, blocking any attempt at standard trigonometric substitution. But fear not—every fortress has a weak point, and in the world of definite integrals, that weak point is often the King's Property.

The King's Gambit

The King's Property states that ${\int }_0^{a}g(x)dx={\int }_0^{a}g(a-x)dx$. When we apply this to our function $f(t)$, we replace every $x$ with $(\pi -x)$.

The numerator transforms from $2x$ to $2(\pi -x)$. Now, look at the denominator: $1-{\cos }^{2}t{\sin }^2(\pi -x)$. Since $\sin (\pi -x)=\sin x$, the denominator remains completely unchanged.

This is our golden opportunity. By adding the original integral to this new version, the $2x$ and the $-2x$ cancel out perfectly:

Dividing by two, we arrive at a much cleaner expression:

The troublemaker is gone.

Symmetry and the Trigonometric Bridge

We have an upper limit of $\pi$, but we know that the function is symmetric around $\pi /2$. Using the property ${\int }_0^{2a}g(x)dx=2{\int }_0^{a}g(x)dx$ (provided $g(2a-x)=g(x)$), we can halve our limit and pull out a factor of two.

Now we have:

To solve this, we need to bring in the heavy artillery: dividing the numerator and denominator by ${\cos }^{2}x$. This transforms our integral into:

By using the identity ${\sec }^{2}x=1+{\tan }^{2}x$, the denominator simplifies to $1+{\tan }^{2}x-{\cos }^{2}t{\tan }^{2}x$, which factors beautifully into $1+{\tan }^{2}x(1-{\cos }^{2}t)$. Since $1-{\cos }^{2}t={\sin }^{2}t$, our denominator is simply $1+{\sin }^{2}t{\tan }^{2}x$.

The Final Transformation

Now, we execute the substitution $z=\tan x$. Consequently, $dz={\sec }^{2}xdx$. As $x$ goes from $0$ to $\pi /2$, $z$ goes from $0$ to $\infty$.

Our integral becomes:

This is a standard integral form: $\int \frac{1}{1+(az{)}^2}dz=\frac{1}{a}{\tan }^{-1}(az)$. Here, $a=\sin t$. Evaluating this from $0$ to $\infty$:

Since ${\tan }^{-1}(\infty )=\pi /2$, the result is:

The Payoff

We have conquered the monster. Now, we return to the final target: ${\int }_0^{\pi /2}\frac{{\pi }^{2}dt}{f(t)}$.

Substituting our hard-won value for $f(t)$, we get:

The antiderivative of $\sin t$ is $-\cos t$. Evaluating from $0$ to $\pi /2$:

The area under the curve is exactly 1. You have navigated the complexity and arrived at a beautiful, singular truth. Well done.