Solve Integral Using Substitution: Integral Question with Example

[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

 

[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}?

 

[Ravenatic20]

You set it up right.

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

 

[rock.freak667]

You set it up right.

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

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

 

[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?

 

[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

 

[Ravenatic20]

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

 

[rock.freak667]

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

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

 

[mrandersdk]

yes, always remeber the new boundaries ;-)