Answer
$\int_{0}^{1}v^2 cos (v^3)~dv = \frac{sin~1}{3}$
Work Step by Step
$\int_{0}^{1}v^2 cos (v^3)~dv$
Let $u = v^3$
$\frac{du}{dv} = 3v^2$
$dv = \frac{du}{3v^2}$
When $v = 0$, then $u = 0$
When $v = 1$, then $u = 1$
$\int_{0}^{1} v^2 cos~u~\frac{du}{3v^2}$
$=\int_{0}^{1} \frac{1}{3} cos~u~du$
$=\frac{1}{3}(sin~u)~\vert_{0}^{1}$
$=\frac{1}{3}(sin~1-sin~0)$
$=\frac{1}{3}(sin~1-0)$
$=\frac{sin~1}{3}$
