How can I simply integrate the expression $\sin(y^2)$ I have 
$$\int_0^1\int_{2x}^2\sin\left(y^2\right)\,dy\,dx$$
and I'm not sure how I can integrate the inner integral without it becoming a monster of a equation that would be overly big to carry on with. Supposed to be done with a function calculator. 
 A: Hint: The bounds of your integral are 
$$
2x\leq y \leq 2\\
0\leq x\leq 1
$$
which is equivalent to saying
$$
0\leq x \leq \frac{y}{2}\\
0\leq y\leq 2
$$
More than a hint:
$$
\int_0^1\int_{2x}^2\sin\left(y^2\right)\,dy\,dx\stackrel{\text{Fubini's}}{=}
\int_0^2\int_{0}^{\frac{y}{2}}\sin\left(y^2\right)\,dx\,dy\\
=\frac{1}{2}\int_0^2y\sin\left(y^2\right)\,dy\\
=\frac{1}{4}\int_0^4\sin\left(u\right)\,du\\
=\frac{1}{4}(1-\cos 4)\\
$$
A: Be
$$\Theta(x)=\begin{cases}
0 & x<0\\
1 & x\ge 0
\end{cases}$$
Then you can rewrite your integral as
\begin{align}
\int_0^1\int_{2x}^2\sin\left(y^2\right)\,\mathrm dy\,\mathrm dx
&= \int_0^1\int_0^2\Theta(y-2x)\sin\left(y^2\right)\,\mathrm dy\,\mathrm dx\\
&= \int_0^2\int_0^1\Theta(y-2x)\sin\left(y^2\right)\,\mathrm dx\,\mathrm dy\\
&= \int_0^2\int_0^1\Theta(y-2x)\,\mathrm dx\,\sin\left(y^2\right)\,\mathrm dy\\
&= \int_0^2\int_0^{y/2}\,\mathrm dx\,\sin\left(y^2\right)\,\mathrm dy\\
&= \int_0^2 \frac{y}{2}\sin\left(y^2\right)\,\mathrm dy\\
\end{align}
The step where $\Theta$ is introduced works because the theta function is $1$ on the original range of the integral and $0$ on the additional range. The removal of the theta function works in an analogous way.
