This is a homework problem, so please do not give more than hints. I must convert \begin{align} \int_0^\sqrt{2}\int_x^\sqrt{4-x^2}\sin\left(x^2+y^2\right)\:dy\:dx\tag{1} \end{align} to polar coordinates. This is my attempt: \begin{align} \int_{\pi/4}^{\pi/2}\int_{\color{red}{2\cos\left(\theta\right)}}^{\color{red}{2\sin\left(\theta\right)}}\sin\left(r^2\right)r\:dr\:d\theta,\tag{2} \end{align} but I am unsure about the $\color{red}{\text{red}}$ limits, because while I am solving I end up at \begin{align} \int_{\pi/4}^{\pi/2}\frac{\cos\left(4\cos^2\left(\theta\right)\right)}{2}-\frac{\cos\left(4\sin^2\left(\theta\right)\right)}{2}\:d\theta\tag{3} \end{align} after a single round of $u$-substitution. There's no way it should end up here, unless it's really easy and I'm just not thinking...
I think the upper limit is $\color{red}{2\sin\left(\theta\right)}$ because a substitution of $2\cos\left(\theta\right)$ into $\sqrt{4-x^2}$ results in \begin{align} \sqrt{4-x^2}&=\sqrt{4-4\cos^2\left(\theta\right)}\\ &=2\sin\left(\theta\right), \end{align} and the lower limit is $\color{red}{2\cos\left(\theta\right)}$ by direct substitution as before.
Here is my $u$-substitution:
Let $\xi=r^2$, then $d\xi/2r=dr$, resulting in \begin{align} ;\;\int r\sin\left(r^2\right)\:dr&=\frac{1}{2}\int \sin\left(\xi\right)\:d\xi\\ &=\frac{-\cos\left(\xi\right)}{2}=\frac{-\cos\left(r^2\right)}{2}\bigg| \end{align} Thus, \begin{align} \int_{\pi/4}^{\pi/2}\int_{2\cos\left(\theta\right)}^{2\sin\left(\theta\right)}r\sin\left(r^2\right)\:dr\:d\theta&=\int_{\pi/4}^{\pi/2}\left[\frac{-\cos\left(r^2\right)}{2}\right]_{2\cos\left(\theta\right)}^{2\sin\left(\theta\right)}\;d\theta. \end{align} Where have I gone wrong?
firstly, you must sketch the region
$$\int_{\pi /4}^{\pi /2}\int_{0}^{2}\sin(r^2)rdrd\theta $$