Answer
$$\frac{{\sqrt 2 }}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\int_0^1 {\int_0^{{x^2}} {x\cos y} dzdxdy} } \cr
& = \int_0^{\pi /4} {\int_0^1 {\left[ {\int_0^{{x^2}} {x\cos y} dz} \right]dxdy} } \cr
& {\text{Solve the inner integral}}{\text{, integrate with respect to }}z,{\text{ treat }}y{\text{ and }}x{\text{ as constants}} \cr
& \int_0^{{x^2}} {x\cos y} dz = \left[ {xz\cos y} \right]_0^{{x^2}} \cr
& {\text{Evaluate the limits in the variable }}z \cr
& = x\left( {{x^2}} \right)\cos y - x\left( 0 \right)\cos y \cr
& = {x^3}\cos y \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^{\pi /4} {\int_0^1 {\left[ {\int_0^{{x^2}} {x\cos y} dz} \right]dxdy} } = \int_0^{\pi /4} {\int_0^1 {{x^3}\cos ydxdy} } \cr
& = \int_0^{\pi /4} {\left( {\int_0^1 {{x^3}\cos ydx} } \right)} dy \cr
& {\text{Integrate with respect to }}x,{\text{ treat }}y{\text{ as a constant}} \cr
& \int_0^1 {{x^3}\cos ydx} = \left( {\frac{{{x^4}}}{4}\cos y} \right)_0^1 \cr
& = \frac{{{{\left( 1 \right)}^4}}}{4}\cos y - \frac{{{{\left( 0 \right)}^4}}}{4}\cos y \cr
& = \frac{1}{4}\cos y \cr
& \int_0^{\pi /4} {\left( {\int_0^1 {{x^3}\cos ydx} } \right)} dy = \frac{1}{4}\int_0^{\pi /4} {\cos y} dy \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}\left( {\sin y} \right)_0^{\pi /4} \cr
& {\text{Evaluate}} \cr
& = \frac{1}{4}\left( {\sin \frac{\pi }{4} - \sin 0} \right) \cr
& = \frac{1}{4}\left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& = \frac{{\sqrt 2 }}{8} \cr} $$
