Prove another identity of the integral of some trig functions

Prove the following identity:
$\int_0^{\pi} x f(\sin x) \, dx = \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx.$
Using part (a) deduce the formula
$\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \pi \int_0^1 \frac{dx}{1+x^2}.$

Proof. Following the hint, we make the substitution
$u = \pi - x \qquad du = -dx.$

So we then have

$\begin{align*} \int_0^{\pi} x f(\sin x) \, dx &= - \int_{u(0)}^{u(\pi)} (\pi - u) f(\sin (\pi - u)) \, du \\ &= -\int_{\pi}^0 (\pi - u) f(\sin u) \, du \\ &= \int_0^{\pi} (\pi - u) f(\sin u) \, du \\ &= \int_0^{\pi} \pi f(\sin u) \, du - \int_0^{\pi} u f (\sin u) \, du \end{align*}$

Here, we change the the name of the variable of integration from $u$ to $x$ . (We can always rename the variable of integration since integrating $f(t) \, dt$ is the same as integrating $f(x) \, dx$ , for example.) So this means we have

$\begin{align*} &&\int_0^{\pi} x f(\sin x) \, dx &= \int_0^{\pi} \pi f(\sin x) \, dx - \int_0^{\pi} x f (\sin x) \, dx \\ \implies && \int_0^{\pi} x f(\sin x) \, dx &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx. \qquad \blacksquare \end{align*}$
Now to deduce the requested formula, first we use the trig identity $\cos^2 x + \sin^2x = 1$ , which implies $1 + \cos^2 x = 2 - \sin^2 x$ , to rewrite the integral
$\begin{align*} \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx &= \int_0^{\pi} \frac{x \sin x}{2 - \sin^2 x} \, dx \\ &= \int_0^{\pi} x f( \sin x) \, dx \qquad \text{where} \qquad f(x) = \frac{x}{2-x^2}. \end{align*}$

Then, using the formula we established in part (a) we have

$\begin{align*} \int_0^{\pi} x f(\sin x) \, dx &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx \\ &= \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{2-\sin^2 x} \, dx \\ &= \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \, dx. \end{align*}$

Now, we use the substitution method, letting

$u = \cos x \qquad du = -\sin x \, dx.$

So we obtain,

$\begin{align*} \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \,dx &= -\frac{\pi}{2} \int_1^{-1} \frac{1}{1+u^2} \, du \\ &= \frac{\pi}{2} \int_{-1}^1 \frac{1}{1+x^2} \, dx &(\text{renaming variable of integration}) \\ &= \pi \int_0^1 \frac{1}{1+x^2} \, dx. \end{align*}$

The last equality follows since $\frac{1}{1+(-x)^2} = \frac{1}{1+x^2}$ so that this is an even function. Hence (by a previous exercise) the integral from -1 to 1 is twice the integral from 0 to $1. \qquad \blacksquare$