Answer
$$\frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt \pi }^{\sqrt {2\pi } } {\int_0^{{x^3}} {\sin \frac{y}{x}} dydx} \cr
& = \int_{\sqrt \pi }^{\sqrt {2\pi } } {\left[ {\int_0^{{x^3}} {\sin \frac{y}{x}} dy} \right]dx} \cr
& {\text{solve the inner integral and treat }}x{\text{ as a constant}} \cr
& = \int_0^{{x^3}} {\sin \frac{y}{x}} dy \cr
& = \left( { - x\cos \frac{y}{x}} \right)_0^{{x^3}} \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left( { - x\cos \frac{{{x^3}}}{x}} \right) - \left( { - x\cos \frac{0}{x}} \right) \cr
& {\text{simplifying}} \cr
& = \left( { - x\cos {x^2}} \right) - \left( { - x\cos 0} \right) \cr
& = - x\cos {x^2} + x \cr
& {\text{then}} \cr
& \int_{\sqrt \pi }^{\sqrt {2\pi } } {\left[ {\int_0^{{x^3}} {\sin \frac{y}{x}} dy} \right]dx} = \int_{\sqrt \pi }^{\sqrt {2\pi } } {\left( { - x\cos {x^2} + x} \right)dx} \cr
& = - \frac{1}{2}\int_{\sqrt \pi }^{\sqrt {2\pi } } {\left( {2x} \right)\cos {x^2}dx} + \int_{\sqrt \pi }^{\sqrt {2\pi } } {xdx} \cr
& {\text{integrating}} \cr
& = \left( { - \frac{1}{2}\sin {x^2} + \frac{{{x^2}}}{2}} \right)_{\sqrt \pi }^{\sqrt {2\pi } } \cr
& = \left( { - \frac{1}{2}\sin {{\left( {\sqrt {2\pi } } \right)}^2} + \frac{{{{\left( {\sqrt {2\pi } } \right)}^2}}}{2}} \right) - \left( { - \frac{1}{2}\sin {{\left( {\sqrt \pi } \right)}^2} + \frac{{{{\left( {\sqrt \pi } \right)}^2}}}{2}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {0 + \pi } \right) - \left( { - 0 + \frac{\pi }{2}} \right) \cr
& = \pi - \frac{\pi }{2} \cr
& = \frac{\pi }{2} \cr} $$
