# Double integration using trig term

1. Jul 22, 2008

### glog

1. The problem statement, all variables and given/known data

$$\int^1_0 \int^1_y sin(x^2) dx dy$$

3. The attempt at a solution

This equation cannot be integrated nicely, so I tried to reverse the order of integration:

$$\int^1_0 \int^1_x sin(x^2) dy dx$$

However this only helps for the first step, since when we intregrate by y, we get:

$$\int^1_0 sin(x^2)-xsin(x^2) dx$$

I'm stuck... again! Any ideas?

2. Jul 22, 2008

### Defennder

Your limits in the new double integral is incorrect. If in doubt, look at the original double integral and draw a picture of the region enclosed by the limits. Then try to express the limits of the double integral in the reversed order of the same region.

3. Jul 22, 2008

### glog

Alright so perhaps the bounds are then: $$\int^1_0 \int^x_0$$

In which case, my integral simplifies to:

$$\int^1_0 x sin(x^2)$$

Which becomes:

-1/2 cos (x^2) | 1_0

This make sense?

4. Jul 22, 2008

### Defennder

Yep, you got it.