Encontrar $I\ =\ \displaystyle\int \dfrac{(2r - 1)\cos \sqrt{3(2r - 1)^2 + 6}}{\sqrt{3(2r - 1)^2 + 6}} dr$.

Encontrar $I\ =\ \displaystyle\int \dfrac{(2r - 1)\cos \sqrt{3(2r - 1)^2 + 6}}{\sqrt{3(2r - 1)^2 + 6}} dr$.

Resolução:

Seja $u = 2r - 1$, $du = 2dr$.

$I\ =\ \dfrac{1}{2} \displaystyle\int \dfrac{u\cos \sqrt{3u^2 + 6}}{\sqrt{3u^2 + 6}} du$.

Seja $v = 3u^2 + 6$, $dv = 6u du$.

$I\ =\ \dfrac{1}{12} \displaystyle\int \dfrac{\cos \sqrt{v}}{\sqrt{v}} dv$

Seja $w = \sqrt{v}$, $dw = \dfrac{dv}{2\sqrt{v}}$.

$I\ =\ \dfrac{1}{6} \displaystyle\int \cos w\ dw\ =\ \dfrac{\sin w}{6} + c\ =\ \dfrac{\sin \sqrt{v}}{6} + c\ =\ \dfrac{\sin \sqrt{3u^2 + 6}}{6} + c$

Logo, $\fbox{$\displaystyle\int \dfrac{(2r - 1)\cos \sqrt{3(2r - 1)^2 + 6}}{\sqrt{3(2r - 1)^2 + 6}} dr\ =\ \dfrac{\sin \sqrt{3(2r - 1)^2 + 6}}{6} + c$}$.