1. ## Two definite integrals

I'm a bit stuck on these. A nudge in the right direction would be much appreciated.

$\int_{- \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {x^{2} sin{x}}{1+x^{6}} dx
$

I can't seem to figure out what to do with that sin(x) function. If I let $u = x^{3}$, I can get the whole thing into the form of the derivative of arctan(x), except for the sin(x).

The other one that's bugging me is:
$\int_{0}^{4} \frac {x}{\sqrt{1+2x}} dx$

No excuse here. I've tried quite a few different substitutions and algebraic manipulations without success. I can't shake the feeling that I'm over-looking something glaring with this problem. I think I found a method that would involve using partial fractions, but I was told that none of these problems required that method.

2. Originally Posted by DarrenM
I'm a bit stuck on these. A nudge in the right direction would be much appreciated.

$\int_{- \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {x^{2} sin{x}}{1+x^{6}} dx
$

I can't seem to figure out what to do with that sin(x) function. If I let $u = x^{3}$, I can get the whole thing into the form of the derivative of arctan(x), except for the sin(x).

The other one that's bugging me is:
$\int_{0}^{4} \frac {x}{\sqrt{1+2x}} dx$

No excuse here. I've tried quite a few different substitutions and algebraic manipulations without success. I can't shake the feeling that I'm over-looking something glaring with this problem. I think I found a method that would involve using partial fractions, but I was told that none of these problems required that method.
1.
The integrated function is odd and the integral is from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$ , What do you think ?

2. The substitution $u=1+2x$ will make it looks like :
$\frac{1}{4}\int_1^9 \frac{u-1}{\sqrt{u}} du$.
Which is easier.

The substitution $u=\sqrt{1+2x}$ works also.

3. Thanks!

1. Zero. I wasn't certain, looking at the function, whether it was even or odd. I know that sin(x) is an odd function, but I wasn't sure about the rest of it. $x^{2}$ is an even function, and $\frac {1}{1+x^{6}}$ is an even function.

I just did a quick search, however, and I think I may have found the justification. The product of an even function and an odd function is odd. The quotient of an odd function and an even function is an odd function. So, despite the fact that both of those were even, the overall result is still odd... is that correct?

2. Argh. I got to that point but didn't actually factor out the 2, hence the "partial fractions" thing. Always a bit embarassed when I make a dumb mistake like that.

Thank you very much for your help.

4. It's worth to say that $\int_{-a}^a f=0$ for $a>0$ when $f$ is odd, as long as this one is integrable.

For example, one could say $\int_{-1}^1\frac{dx}x=0,$ but this function is not bounded on $[-1,1]$ so it's not integrable there and the result does not apply.