# Double integration of trig function

The function I am trying to integrate is:

$$\int_0^{12}\int_{y/4}^3 \sin(x^2)\ \mathrm{d}y\ \mathrm{d}x$$

I am having trouble finding out where to start. I have tried to just go ahead and integrate, but then I would have to deal with calculating cosine of $9$ and $\left(\frac{y}{4}\right)^3$. I thought about changing the order of integration, but how would you do that considering the inner bound is a function of $x$?

I would greatly appreciate some help to point me in the right direction.

• Draw a picture of your region. That will help you determine the bounds for your change of variable. You should find swictching the order gives $$\int_{0}^{3} \int_{0}^{4x} \sin(x^{2}) dydx$$ Also, I think your integration should go $dxdy$ not $dydx$ as you have it now. – mattos Apr 19 '18 at 1:26
• You can't just "go ahead in integrate" $\int \sin (x^2) dx$ does not integrate into elemetary functions. But if you change the order of integration, then a substitution will make itself available to you such that you will have something that is integrable. – Doug M Apr 19 '18 at 1:28
• @Mattos Thank you for your concise explanation. I also thought it would work only as dxdy, not dydx. I'm assuming my professor made a typo on the worksheet. – user78100 Apr 19 '18 at 1:38
• @user78100 The inner integral as you have it is asking you to integrate a function with bounds in $y$ with respect to $y$, so either there is a typo or the professor is doing something I've never heard of. – mattos Apr 19 '18 at 1:44
• I second @Mattos first comment. That’s why I prefer the notation $$\int_0^{12}\mathrm{d}y\int_{y/4}^3\mathrm{d}x\sin x^2$$ – gen-z ready to perish Apr 19 '18 at 4:46

\begin{align} I&=\int_0^{12}\left( \int_{y/4}^{3}\sin (x^2)dx\right)dy\\&=\int_0^3dx\int_0^{4x} \sin(x^2)dy\\&=\int_0^3\sin(x^2)4xdx \\ &= 2\int_0^3\sin(x^2)d(x^2)\\&=2\int_0^9\sin udu \\&=2(1-\cos 9) \end{align}