Evaluate integral from 0 to pi of (x*sin(x)) / (1 + $cos^2(x)$) dx and show that it equals $pi^2 / 4$.

Step 1: Apply Property of Definite Integrals

We use the property of definite integrals that states $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$. Here, $a = \pi$. We substitute $x$ with $(\pi - x)$ in the integrand. Note that $\sin(\pi - x) = \sin(x)$ and $\cos(\pi - x) = -\cos(x)$, so $\cos^2(\pi - x) = (-\cos(x))^2 = \cos^2(x)$.

Step 2: Simplify and Add the Integrals

Now we have two expressions for $I$. We add the original integral and the transformed integral. The denominators are the same, so we can combine the numerators. The $x$ terms cancel out, leaving $\pi \sin(x)$ in the numerator.

Step 3: Substitute to Simplify the Integral

To evaluate the integral, we use a substitution. Let $t = \cos(x)$. Then, $dt = -\sin(x) dx$. We also need to change the limits of integration. When $x = 0$, $t = \cos(0) = 1$. When $x = \pi$, $t = \cos(\pi) = -1$.

Step 4: Evaluate the Integral

We can change the order of integration by negating the integral: $\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$. The integral of $\frac{1}{1 + t^2}$ is $\tan^{-1}(t)$. We evaluate this at the new limits, $1$ and $-1$. We know that $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(-1) = -\frac{\pi}{4}$.

Step 5: Solve for I

Finally, we divide both sides by 2 to find the value of I.

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