(i)
(ii) Since cos ( x 3 ) {\displaystyle \displaystyle \cos(x^{3})} is hard to integrate with respect to x {\displaystyle \displaystyle x} , we change the order of integration. Using the sketch of (i) and that x = − y {\displaystyle \displaystyle x={\sqrt {-y}}} is equivalent to y = − x 2 {\displaystyle \displaystyle y=-x^{2}} (and x ≥ 0 {\displaystyle \displaystyle x\geq 0} ) we find that
∫ − 4 0 ∫ − y 2 cos ( x 3 ) d x d y = ∫ 0 2 ∫ − x 2 0 cos ( x 3 ) d y d x = ∫ 0 2 cos ( x 3 ) [ y ] − x 2 0 d x = ∫ 0 2 cos ( x 3 ) x 2 d x = [ 1 3 sin ( x 3 ) ] 0 2 = 1 3 sin ( 8 ) − 1 3 sin ( 0 ) = 1 3 sin ( 8 ) {\displaystyle \displaystyle {\begin{aligned}\int _{-4}^{0}\int _{\sqrt {-y}}^{2}\cos(x^{3})\;{\text{d}}x{\text{d}}y&=\int _{0}^{2}\int _{-x^{2}}^{0}\cos(x^{3})\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{2}\cos(x^{3})[y]_{-x^{2}}^{0}\;{\text{d}}x\\&=\int _{0}^{2}\cos(x^{3})x^{2}\;{\text{d}}x\\&=[{\frac {1}{3}}\sin(x^{3})]_{0}^{2}={\frac {1}{3}}\sin(8)-{\frac {1}{3}}\sin(0)\\&={\frac {1}{3}}\sin(8)\end{aligned}}}
![{\displaystyle \displaystyle {\begin{aligned}\int _{-4}^{0}\int _{\sqrt {-y}}^{2}\cos(x^{3})\;{\text{d}}x{\text{d}}y&=\int _{0}^{2}\int _{-x^{2}}^{0}\cos(x^{3})\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{2}\cos(x^{3})[y]_{-x^{2}}^{0}\;{\text{d}}x\\&=\int _{0}^{2}\cos(x^{3})x^{2}\;{\text{d}}x\\&=[{\frac {1}{3}}\sin(x^{3})]_{0}^{2}={\frac {1}{3}}\sin(8)-{\frac {1}{3}}\sin(0)\\&={\frac {1}{3}}\sin(8)\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0623d08715d3f6bdb0d71ac671e390a69c63ded2)