Evaluate $\int_0^3\sqrt\frac{x^3}{3-x}dx$. 
Evaluate $\int_0^3\sqrt\frac{x^3}{3-x}dx$.

We can solve it by putting $x=3\sin^2\theta$. But it's not really an intuitive method.
Is there any other way to solve it?
Or, is there some pattern to this question that tells us instinctively that the substitution is $x=3\sin^2\theta$?

You can do the substitution $t^2=\frac{x}{3-x}$, since $$\sqrt{\frac{x^3}{3-x}}=x\sqrt{\frac{x}{3-x}}\qquad\text{for }x\ge0.$$
