Having trouble with indefinite integration by substitution. See problem below:

Question

anonymous · Answer

$$\int\limits \sqrt{t^3-1} t^2 dt$$
I understand that
$$u=t^3-1$$ and that$$du = 3t^2 dt$$

anonymous · Answer

Where I get confused is this. How am I to get to $$\frac{1}{3} \int\limits \sqrt{u} du$$

anonymous · Answer

Solve for dt and plug both equations back into the first equation. the t^2 will cancel out, and since 1/3 is a constant, it can be taken out of the integral.

anonymous · Answer

What steps am I supposed to take to solve for dt? (Sorry, it's probably blatantly obvious to you).

anonymous · Answer

if $$du = 3t ^{2} dt$$
then divide both sides by 3t^2 to get:
$$du/3t ^{2} = dt$$

anonymous · Answer

AAAH! Ok, I'm going to give that a try...

anonymous · Answer

$$du = 3t^{2}dt$$ 
re-write this equation in terms of dt.
Then plug in $$dt = du/(3t^{2})$$ in your indefinite integral and voila you have an intergral in terms of du.  
That's what gets the equation you asked about
$$1/3\int\limits_{?}^{?}\sqrt{u}du$$

anonymous · Answer

Thank you both, from that I get$$\frac{2}{9}(t^3-1)^{\frac{3}{2}}+C$$