Integrate $\int_{0}^{2}{\int_{y^2}^{4}{y\cos(x^2)dxdy}}$ I'm asked to evaluate $\int_{0}^{2}{\int_{y^2}^{4}{y\cos (x^2)dxdy}}$. Letting $f(x)=\cos (x^2)$ We have have that $\int_{y^2}^{4}{\cos (x^2)dx} = F(4)-F(y^2)$ by the FTC. This gives us
\begin{align*}
\int_{0}^{2}{\int_{y^2}^{4}{y\cos(x^2)dxdy}} &= \int_{0}^{2}{(yF(4)-yF(y^2))dy}\\
&=2F(4)-\int_{0}^{2}yF(y^2)dy\\
&=2F(4)-(\frac{y^2}{2}F(4) \big|_0^2 - \int_{0}^{2}{\frac{y^2}{2} \cdot 2y \cdot f(y^2)dy})\\
&= \int_{0}^{2}{y^3\cos(y^4)dy} = \frac{\sin(16)}{4}
\end{align*}
Is this a valid way to evaluate this integral? The solution given to me changed the order of integration instead and did it a bit differently. Thanks.

hint: Switch the order is better: $I = \int_{0}^ 4\int_{0}^{\sqrt{x}} y\cos(x^2)dydx$
