Calcular $I\ =\ \displaystyle\int_0^\pi \sin^2 \left(1 + \dfrac{\theta}{2}\right)\ d\theta$.

$I\ =\ \displaystyle\int_0^\pi \cos^2 \left(\dfrac{\pi}{2} - 1 - \dfrac{\theta}{2}\right)\ d\theta\ =\ \displaystyle\int_0^\pi \dfrac{\cos (\pi - 2 - \theta) + 1}{2}\ d\theta$

 

Seja $u = \pi - 2 - \theta$, $du = -d\theta$.

 

$I\ =\ \displaystyle\int_{-2}^{\pi - 2} \dfrac{1 + \cos u}{2}\ du\ =\ \left.\dfrac{u}{2}\right|_{-2}^{\pi - 2} + \left.\dfrac{\sin(u)}{2}\right|_{-2}^{\pi - 2} =$

 

$= \dfrac{\pi - 2}{2} + 1 + \dfrac{\sin(\pi - 2)}{2} + \dfrac{\sin 2}{2} = \fbox{$\dfrac{\pi}{2} + \sin 2$}$