A Difficult Definite Integral $\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$ Problem
Evaluate $$\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$$
Comment
It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula

$$\int_0^{2\pi}xf(\cos x){\rm d}x=\pi\int_0^{2\pi}f(\sin x){\rm d}x,$$
where $f(x) \in C[-1,1].$

\begin{align*}
\require{begingroup}
\begingroup
\newcommand{\dd}{\;{\rm d}}\int_0^{2\pi} (t-\sin t)(1-\cos t)^2 \dd t
&= \int_0^{2\pi} t(1-\cos t)^2 \dd t - \int_0^{2\pi} \sin t(1-\cos t)^2 \dd t \\
&= \pi\int_0^{2\pi} (1-\sin t)^2 \dd t - \int_0^{2\pi} (1-\cos t)^2 \dd (1-\cos t) \\
&= \pi\int_0^{2\pi} \left(\frac32-\frac12\cos2t-2\sin t\right) \dd t - \left[\frac13(1-\cos t)^3\right]_0^{2\pi}\\
&= \pi\left[\frac32t-\frac14\sin2t+2\cos t\right]_0^{2\pi}\\
&= 3\pi^2
\endgroup
\end{align*}
But any other solution?

\begin{align}
&\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t=\int_0^{2\pi}t+t{\cos }^2 t-2t\cos t-\sin t -\sin t{\cos }^2 t+\sin 2t{\rm d}t\\
&=\left[\frac{1}{2}t^2+\frac{1}{2}t^2+\frac{1}{4}t\sin 2t-\frac{1}{4}t^2+\frac{1}{8}\cos 2t-2t\sin t-2\cos t+\cos t+\frac{1}{3}{\cos }^3t-\frac{1}{2}\cos 2t\right]_0^{2\pi} \\
&=\left[\frac{3}{4}t^2+\frac{1}{4}t\sin 2t-\frac{3}{8}\cos 2t-2t\sin t-\cos t+\frac{1}{3}{\cos }^3 t\right]_0^{2\pi}\\
&=3\pi ^2
\end{align}
