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 $x=3\sin^2\theta$?
 A: 
Or, is there some pattern to this question that tells us instinctively that the substitution is $x=3\sin^2\theta$?

To the familiar eye, yes, there is.
Note powers of $x,\,3-x$ are important here. Whenever two quantities in the integral you should care about have a nonzero constant sum, it's worth leveraging $\sin^2\theta+\cos^2\theta=1$, or possibly another sum-of-squares identity such as $\tanh^2\phi+\operatorname{sech}^2\phi=1$, but with the right transformation these are basically the same idea.
Similarly, if two quantities you have to care about have a nonzero constant difference, exploit $\sec^2\theta-\tan^2\theta=1$ or, by the same logic, maybe $\cosh^2\phi-\sinh^2\phi=1$.
(Mind you, out of circular vs hyperbolic you often find one more directly helpful with whichever problem is at hand, e.g. I'd recommend the circular option here.)
Once you've switched to trigonometry, further tips often write the rest of the calculation for you. Actually, in this case the latter brings you back round to no trigonometry at all; the real trick is $x=3y$. But diving into trigonometry often helps you spot a useful rational transformation, which may well be much less trivial in other cases.
A: You can do the substitution $t^2=\frac{x}{3-x}$, since $$\sqrt{\frac{x^3}{3-x}}=x\sqrt{\frac{x}{3-x}}\qquad\text{for }x\ge0.$$
