# Evaluate the integral

• Apr 6th 2013, 08:34 AM
vorb
Evaluate the integral
Attachment 27831
This was an exam question from a previous year, but I'm having a lot of trouble just finding the antiderivative of this.
I've tried factoring the bottom since it's a sum of cubes, and afterwards attempting to break it into partial fractions, but it just ends up being an insane mess (Headbang)
Could someone just steer me in the right direction of approaching this? I've been going at this for the past hour to no avail.
• Apr 6th 2013, 08:43 AM
MINOANMAN
Re: Evaluate the integral
Choose A. : π/48
• Apr 6th 2013, 08:45 AM
vorb
Re: Evaluate the integral
Quote:

Originally Posted by MINOANMAN
Choose A. : π/48

I know what the answer is -__-"
I wanted help on approaching how to solve something like this....

As on any math forum, giving THE answer to a question never really helps anyone...
• Apr 6th 2013, 09:05 AM
MINOANMAN
Re: Evaluate the integral
use simpsons method or trapezium rule....and evaluate this integral
Simpson's rule - Wikipedia, the free encyclopedia
Trapezoidal rule - Wikipedia, the free encyclopedia
• Apr 6th 2013, 09:19 AM
HallsofIvy
Re: Evaluate the integral
Do each part separately: $\int_{-2}^2 \frac{x^5dx}{x^6+ 64}+ \int_{-2}^2\frac{x^2 dx}{x^6+ 64}+\int_{-2}^2 \frac{4xdx}{x^6+ 64}+ \int_{-2}^2\fra{sin(x) dx}{x^6+ 64}$

In the first one, let $u= x^6+ 64$. In the second, let $u= x^3$. In the third, let $u= x^2$. The last is trivial because the integrand is an odd function.
• Apr 6th 2013, 09:33 AM
vorb
Re: Evaluate the integral
Quote:

Originally Posted by HallsofIvy
Do each part separately: $\int_{-2}^2 \frac{x^5dx}{x^6+ 64}+ \int_{-2}^2\frac{x^2 dx}{x^6+ 64}+\int_{-2}^2 \frac{4xdx}{x^6+ 64}+ \int_{-2}^2\fra{sin(x) dx}{x^6+ 64}$

In the first one, let $u= x^6+ 64$. In the second, let $u= x^3$. In the third, let $u= x^2$. The last is trivial because the integrand is an odd function.

Thanks! Just what I was looking for!
• Apr 6th 2013, 11:18 AM
hollywood
Re: Evaluate the integral
Quote:

Originally Posted by HallsofIvy
Do each part separately: $\int_{-2}^2 \frac{x^5dx}{x^6+ 64}+ \int_{-2}^2\frac{x^2 dx}{x^6+ 64}+\int_{-2}^2 \frac{4xdx}{x^6+ 64}+ \int_{-2}^2\fra{sin(x) dx}{x^6+ 64}$

In the first one, let $u= x^6+ 64$. In the second, let $u= x^3$. In the third, let $u= x^2$. The last is trivial because the integrand is an odd function.

The first and third are also odd functions.

- Hollywood