how can i change specifically the intervals of a double integral? I know how to change the intervals of an integral, for example the integral of $(\sin x)^2$ from $-\pi$ to $\pi$ is equal to $\pi\int_{-1}^1 (\sin πx)^2 \,dx$. 
I find it difficult to do that in 2D. In particular i want the
$$
\int_1^2 \int_3^4 \sin(\pi x)\sin(\pi y)\, dx\, dy
$$
to change intervals and become $(-1,1)$ $(-1,1)$
**Thanks for the information but i would like specifically to do a double integration on the square [-1,1]^2 . I mean that $$
\int_1^2 \int_3^4 \sin(\pi x)\sin(\pi y)\, dx\, dy
$$
i want it to change to $$
\int_{-1}^1 \int_{-1}^1 $$
 A: In this case,
since the integrand
is just a product
of functions of
$x$ and $y$,
you can write it as
$$\int_3^4 \sin(\pi x) dx
\int_1^2 \sin(\pi y) dy
.$$
The linear transformations
 should then be
straightforward.
This assumes that
$dx$ being inside
means that
the inner integral
is on $x$.
A: In higher dimensions,we usually don't simply change intervals by multiplying by a real number,it's usually more complex then that. What we cando,which is more powerful and versatile,is try and transform the region from Cartesian to polar coordinates. In this case,though-not only would the resulting computation, using the addition and product angle formulas,be vastly more complicated,it's not necessary. Because the integrand is a sine function ,which is infinitely differentiable on the region. by the Cauchy-Fubini Theorum, we can rewrite the double integral as a product of integrals: 
$$\int_3^4 \sin(\pi x) dx \int_1^2 \sin(\pi y) dy.$$
Also,by the Cauchy-Fubini Theorem,the order of integration with respect to x or y shouldn't matter. 
