Integrating a non-elementary function

[bosox09]

Homework Statement

Integral (pi^1/2, 0) of Integral (pi^1/2, y) sin(y²) dxdy

Homework Equations

This one is interesting because it can't be integrated as is (at least not at the level of my course) but I think with some rearranging it can be done. I was wondering if anyone could verify my method.

BTW, I'm kinda new, so I don't know exactly how to make the little integral symbol. If that's too hard to read, it's a double integral defined on the inside between "y" and the square root of pi, and defined on the outside between 0 and the square root of pi. The function is sin(y²), with dx, then dy. That is, I believe 0 < y < pi^1/2 and y < x < pi^1/2.

The Attempt at a Solution

Alright so as is, it's not an elementary function and cannot be worked with, I believe. I rearranged the "parameters" to define x as 0 < x < pi^1/2 and 0 < y < x. Then, "dx" and "dy" switch positions in the original equation, and I integrated sin(y²) with respect to "x" first. Simply put, I got xsin(y²). Then, integrating with respect to "y" by parts, I got xsin(y²) - [(1/2)(x²) * -(2y)(cos(y²))]. Without doing all the math here, I'll tell you I ended up with pi².

I feel I got the problem right all the way until the final integration. I'm mostly afraid I messed up the final step in integration. If anyone would like to look at my process, that would be great.

PS -- if anyone knows how to directly integrate a non-elementary function, that would be cool to see.

 

[sutupidmath]

Is your integral:


$$\int_{0}^{\sqrt\pi}\int_{y}^{\sqrt\pi}sin(y^2)dxdy$$


?

If this is the integral you are working with then, notice that you are first integrating with respect to x, so you can treat

$$sin(y^2)$$ merely as a constant. And after you have integrated with respect to x, you will end up with an integral that can be easily integrated with a u-substitution.

 

[bosox09]

Sorry! Good point.

The integral I put was not the original. I rearranged the original to get to that point, so ignore my "process" as that's how I got to the formula above. Thanks.