# Thread: Integral involving e^(-x^2).

1. ## Integral involving e^(-x^2).

I hope that this is the right place to post this.

I'm studying for my exams this summer and on one of the practice papers there is this integral, which I'm not entirely sure how to approach:

$\int{\exp{(-x^2)}\sqrt{x^4 + 1}\sin{(x)}}dx$

Can anybody help me with this?

2. Originally Posted by penumbra
I hope that this is the right place to post this.

I'm studying for my exams this summer and on one of the practice papers there is this integral, which I'm not entirely sure how to approach:

$\int{\exp{(-x^2)}\sqrt{x^4 + 1}\sin{(x)}}dx$

Can anybody help me with this?

was this a originally a definite integral? ... if not, I doubt it can be done.

3. It was, I'm sorry. It was between pi and -pi.

4. Originally Posted by penumbra
It was, I'm sorry. It was between pi and -pi.
Try doing $\int_{-\pi}^{\pi}=\int_{-\pi}^{0}+\int_0^{\pi}$ and making the sub $x=-z$ in the first integral.

5. Originally Posted by penumbra
It was, I'm sorry. It was between pi and -pi.
easy then. the definite integral's value is 0.

now ... why?

6. Originally Posted by Drexel28
Try doing $\int_{-\pi}^{\pi}=\int_{-\pi}^{0}+\int_0^{\pi}$ and making the sub $x=-z$ in the first integral.
Ahh, I see. I think I will be able to do this now. Thank you.

7. Originally Posted by skeeter
easy then. the definite integral's value is 0.

now ... why?
Because sine of 0, pi and -pi is 0?

8. Originally Posted by penumbra
Because sine of 0, pi and -pi is 0?
no.

note that the integrand is an odd function.

9. Originally Posted by skeeter
no.

note that the integrand is an odd function.
So, then $\int^{\pi}_0f(x)dx = -\int^0_{-\pi}f(x)dx$ and so $\int^{\pi}_0f(x)dx + \int^0_{-\pi}f(x)dx = 0$?

10. if $f(x)$ is an odd function, then $\int_{-a}^a f(x) \, dx = 0$

this is the entire point of the question.

11. OK, thank you.