# Integral of Sin(x^2)

1. Oct 8, 2015

### Dewgale

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

Evaluate the integral $$\int_{y=0}^{y=1} \int_{x=y}^{x=1} sin(x^2) \, dx \, dy$$

2. Relevant equations
N/A

3. The attempt at a solution
We know right away that $sin(x^2)$ has no elementary anti-derivative. Therefore, I analyzed the Maclaurin series of $sin(x)$.

$$sin(x)\ =\ x\ -\ \frac{x^3}{3!}\ +\ \frac{x^5}{5!}\ -\ \frac{x^7}{7!}\ +\ \dots$$
$$sin(x)\ =\ \sum_{n=0}^\infty \frac{(-1)^n(x)^{2n+1}}{(2n+1)!}$$
Therefore,
$$sin(x^2)\ =\ \sum_{n=0}^\infty \frac{(-1)^n(x)^{4n+2}}{(2n+1)!}$$
I then substituted this sum into the original integral.
$$\int_{y=0}^{y=1} \int_{x=y}^{x=1} \sum_{n=0}^\infty \frac{(-1)^n(x)^{4n+2}}{(2n+1)!} \, dx \, dy$$

As $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}$ is a constant, we can bring it outside of the integral. Then the integral becomes
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_{y=0}^{y=1} \int_{x=y}^{x=1} x^{4n+2} \, dx \, dy$$
I then integrated with respect to x, and got
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_{y=0}^{y=1} \big( \left. \frac{x^{4n+3}}{4n+3} \right|_y^1 \big) \, dy$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_{y=0}^{y=1} \big( \frac{1^{4n+3}}{4n+3}\ -\ \frac{y^{4n+3}}{4n+3} \big) \, dy$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{1}{4n+3} \int_{y=0}^{y=1} \big(1^{4n+3}\ -\ y^{4n+3} \big) \, dy$$
I then integrated with respect to y.
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{1}{4n+3} \big( (1^{4n+3})\ -\ \frac{1^{4n+4}}{4n+4} \big)$$
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{1^{4n+3}}{4n+3} \ -\ \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{1}{4n+3} \frac{1^{4n+4}}{4n+4}$$

Now here is where I hit a snag. I can brute force calculate using Wolfram|Alpha to show that the first series converges to $\approx \ 0.310268,$ and the second series converges to $\approx \ 0.08042$. Therefore, the integral converges to $0.310268\ -\ 0.08042\ \approx\ 0.2299848$

This seems remarkably inelegant though, and I'm sure the prof isn't ok with the brute forcing. I'm just not sure how to progress beyond this point. Any help is appreciated greatly!!

2. Oct 8, 2015

### axmls

You're over complicating things. Try to flip the order of the integrals.

3. Oct 12, 2015

### Dewgale

I'm confused as to how that will help, as $\int_{y=0}^{y=1}\ sin(x^2) \, dy\ =\ sin(x^2)$, which still doesn't resolve the fact that $sin(x^2)$ has no elementary integral, and will also leave a y in there if I try using the Maclaurin series, which is just even more ugly. I'm clearly missing something, I just don't know what.

4. Oct 12, 2015

### ehild

Draw the integration domain. You have to integrate over the blue area. What are the limits if you integrate with respect to y first, and then with respect to x?

5. Oct 12, 2015

### Dewgale

Ok, thank you so much! I rewrote the limits and managed to solve the integral as $-\frac{1}{2}\ (cos(1)\ -\ 1)$ which when approximated comes out to the same result as my (much, much bulkier) sums method, which was approximately 0.2298.

6. Oct 12, 2015

### ehild

You are welcome. :)

7. May 16, 2016

### Cooloumb

Could you please explain a little more how you switch the order of integration? Thanks!

8. May 16, 2016

### ehild

You have to integrate over the blue triangle, enclosed by the lines y=0, y=x, and x=1. You can integrate by x from x=y to x=1 first:
and then by y from 0 to 1 $\int_{y=0}^{y=1}\left( \int_{x=y}^{x=1} sin(x^2) \, dx \right )\, dy$
or you integrate by y from 0 to y=x, and then by x from 0 to 1:
$\int_{x=0}^{1=1}\left( \int_{y=0}^{y=x} sin(x^2) \, dy \right )\, dx$
The advantage of the second method is that you integrate a constant with respect to y first, so you get x sin2(x) which can be easily integrate with respect to x.