# Integral Question

1. Feb 5, 2008

### Ravenatic20

Need to solve this using substitution:

$$\int$$$$^{0}$$ y$$^{2}$$ (1- $$\frac{y^{3}}{a^{2}}$$)$$^{-2}$$ dy
$$_{-a}$$

This is my first post so I'm not that good with the LaTeX code yet, but I hope you can read it correctly. I was thinking about substituting u for y^3. What do you guys think? Thanks

2. Feb 5, 2008

### rock.freak667

$$\int_{-a} ^{0} y^2(1-\frac{y^3}{a^2})^{-2} dy$$

instead of u=y^3 why not let u=1-$\frac{y^3}{a^2}$?

3. Feb 5, 2008

### Ravenatic20

You set it up right.

Hmm... If y^3 is used, can't that be used later to cancel out y^2?

4. Feb 5, 2008

### rock.freak667

If you let u=y^3 the y^2 will cancel out but you will be left with

$$\int \frac{1}{3}(1-\frac{u}{a^2})^{-2} du$$

5. Feb 5, 2008

### Ravenatic20

Yea that's what I got, so the y's are out. The 1/3 is constant so that comes out. The tricky part here is the ^-2. Ideas?

6. Feb 6, 2008

### mrandersdk

but if you use $$1-\frac{y^3}{a^2}$$ like suggested, you get

$$du = - \frac{y^2}{3 a^2} dx$$

so

$$\int y^2 (1-\frac{y^3}{a^2}) dx =- 3 a^2 \int \frac{1}{u^2} du$$

this is a bit easier

7. Feb 6, 2008

### Ravenatic20

That works too, but you guys are forgetting the -a to 0 part.

8. Feb 6, 2008

### rock.freak667

Putting back the 0 to -a with a u substitution is wrong, since 0 and -a are the limits for y not u.

If the OP wanted to put back limits,this would have to be done

$$u=1-\frac{y^3}{a^2}$$

y=0;u=1
y=-a;u=1+a

so that

$$\int_{-a} ^{0} y^2(1-\frac{y^3}{a^2})^{-2} dy \equiv \frac{-a^2}{3} \int_{1+a} ^{1} u^{-2} du$$

9. Feb 6, 2008

### mrandersdk

yes, always remeber the new boundaries ;-)