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

• April 14th 2010, 11:07 AM
penumbra
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?

• April 14th 2010, 02:35 PM
skeeter
Quote:

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.
• April 14th 2010, 02:37 PM
penumbra
It was, I'm sorry. It was between pi and -pi.
• April 14th 2010, 02:41 PM
Drexel28
Quote:

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.
• April 14th 2010, 02:43 PM
skeeter
Quote:

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?
• April 14th 2010, 02:52 PM
penumbra
Quote:

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.
• April 14th 2010, 03:38 PM
penumbra
Quote:

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

now ... why?

Because sine of 0, pi and -pi is 0?
• April 14th 2010, 03:51 PM
skeeter
Quote:

Originally Posted by penumbra
Because sine of 0, pi and -pi is 0?

no.

note that the integrand is an odd function.
• April 14th 2010, 04:03 PM
penumbra
Quote:

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 $\int^{\pi}_0f(x)dx + \int^0_{-\pi}f(x)dx = 0$?
• April 14th 2010, 04:12 PM
skeeter
if $f(x)$ is an odd function, then $\int_{-a}^a f(x) \, dx = 0$

this is the entire point of the question.
• April 14th 2010, 04:20 PM
penumbra
OK, thank you.