Answer
$$\frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\root 3 \of {\pi /2} } {\int_0^x {y\cos {x^3}dy} dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^{\root 3 \of {\pi /2} } {\left[ {\frac{{{y^2}}}{2}\cos {x^3}} \right]_0^xdx} \cr
& = \int_0^{\root 3 \of {\pi /2} } {\left[ {\frac{{{x^2}}}{2}\cos {x^3} - \frac{{{0^2}}}{2}\cos {x^3}} \right]dx} \cr
& = \frac{1}{2}\int_0^{\root 3 \of {\pi /2} } {{x^2}\cos {x^3}dx} \cr
& = \frac{1}{6}\int_0^{\root 3 \of {\pi /2} } {3{x^2}\cos {x^3}dx} \cr
& {\text{Integrate }} \cr
& = \frac{1}{6}\left[ {\sin {x^3}} \right]_0^{\root 3 \of {\pi /2} } \cr
& {\text{Evaluating}} \cr
& = \frac{1}{6}\left[ {\sin {{\left( {\root 3 \of {\frac{\pi }{2}} } \right)}^3} - \sin {{\left( 0 \right)}^3}} \right] \cr
& = \frac{1}{6}\left[ {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( 0 \right)} \right] \cr
& = \frac{1}{6} \cr} $$
