# Thread: Finding integrals

1. ## Finding integrals

find the indefinite integral of (x^2)/[(9-x^2)^(1/2)]
Im think im supposed to try to set x=3sin(u) thus dx=3cos(u) du i think im on the right track but i just keep getting the integral of 9sin^2(u) du and i get all messed up
i hope that makes sense. i dont know how to type all crazy math like

2. That's very good. Now use the identity: $\sin^2 x = \frac{1 - \cos 2x}{2}$

This is much easier to integrate.

3. okay then i split it up as 2 integrals like so 9{intieral(1/2)-integral[cos(2u)/2]}
then the the antiderivative of those come out to 9[(u/2)-(sin(2u)/4)]
then when i plug arcsin(x/3) into u it gets all messed up and i have to take the sin of the arcsin?
i dont know if im doing something wrong

4. Use the double angle identity for Sine:

$\sin{2x} = 2\sin{x}\cos{x}$

Then I'm sure you know how to figure out $\sin{x}$ and $\cos{x}$

5. umm i understand that the sin(u) would then equal x/3 but what happens to the cos(u)? i have cos(u)=dx/3?

6. $tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

$sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$

$cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$

7. Where did we get a triangle from?

8. $sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}~=~\frac{ x}{3}$

Now using this information find out what $cos\theta$ is.

Hint use pythargeon theorem to find the adjacent side

9. finally i see it thank you!!!