Calcular $I\ =\ \displaystyle\int \sin (\sqrt{x})\ dx$.

$I\ =\ x\sin (\sqrt{x}) - \displaystyle\int \dfrac{x\cos (\sqrt{x})}{2\sqrt{x}}\ dx$

Seja $u = \sqrt{x}$, $du = \dfrac{dx}{2\sqrt{x}}$.

$I\ =\ x\sin (\sqrt{x}) - \displaystyle\int u^2\cos u\ du$

$I\ =\ x\sin (\sqrt{x}) - u^2\sin u + 2\displaystyle\int u\sin u\ du$

$I\ =\ x\sin (\sqrt{x}) - u^2\sin u - 2u\cos u + 2\displaystyle\int \cos u\ du$

$I\ =\ \cancel{x\sin (\sqrt{x})} - \cancel{x\sin (\sqrt{x})} - 2\sqrt{x}\cos (\sqrt{x}) + 2\sin (\sqrt{x}) + c$

$\fbox{$I = -2\sqrt{x}\cos (\sqrt{x}) + 2\sin (\sqrt{x}) + c$}$