1. ## Help solving integral

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}}$

Can anyone give me a hint where to go with this?

2. Originally Posted by paupsers
$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}}$

Can anyone give me a hint where to go with this?
Let $\displaystyle x=\sin\theta$

3. Originally Posted by Sparta935
First, you need to know that anything Square root(a^2 - x^2) can be rewritten as x=a*sin(theta), so replace x with asin(theta) making dx = cos(theta)dtheta, now either squareroot(of cos^2) = +/- cos(theta) depending on the valuevalues taken by theta. If we choose -pi/2<=0<=pi/2 then you take the integral x/cos(theta) and you can now seperate the function into integral x + integral 1/cos(theta), and you'll have your answer.
A few corrections.

4. Originally Posted by Sparta935
Thanks, lol I am viewing this site via ps3, so it's kind of hard to remain accurate, but the calculations still work, this problem is actually shown in my calculus book in an example lol.
Let's wait for the OP to reply. If you like, we can pm to discuss further.

5. Thanks, I only needed that first tip. I'm actually in Adv. Calc (proof calc) but the section we're in now has all these Calc I/II problems! I wasn't sure whether I should look in my old calc book for trig sub, u sub, by parts, etc.

Thanks for the help!

6. Another cool way to do this is let $\displaystyle u=1-x^2$, (which also implies that $\displaystyle x^2=1-u$)
Then $\displaystyle du=-2xdx$

You can then substitute $\displaystyle 1-x^2$ with $\displaystyle u$, $\displaystyle x^2$ with $\displaystyle 1-u$, the "other" $\displaystyle x$ in the denominator with $\displaystyle du$ (as long you put the $\displaystyle -\frac{1}{2}$ out front.) Then the integral becomes $\displaystyle -\frac{1}{2}\int \frac{1-u}{u^\frac{1}{2}}du$, which can then be easily evauluated by bringing up the $\displaystyle u^\frac{1}{2}$ to the denominator