Evaluate the given integral What is the simplest way to evaluate the integral $$\int x(a^2-x^2)^{\frac{1}{2}}\, dx ?$$
Does trigonometric substitution $x=a \sin \theta$ work ? 
 A: The easier route is to substitute $$u = a^2-x^2\implies du = -2x\,dx$$
Then you have $$\begin{align} \int x(a^2 - x^2)^{1/2}\,dx & = -\frac 12\int (\underbrace {a^2 - x^2}_u)^{1/2}\underbrace{(-2x\,dx)}_{du}\\ \\ & = -\frac 12 \int u^{1/2} \, du\\ \\ &= -\frac 12 \frac{u^{3/2}}{3/2} + C \\ \\ & = -\frac 13(a^2 - x^2)^{3/2} + C \end{align}$$ 
If you didn't have the $x$ factor in your integral, i.e. if you had only $$\int  \sqrt{a^2 - x^2}\,dx$$
...then, yes, trigonometric substitution would work nicely.
A: First I do this the quick method, and than I fully explain your method of trig substation and shows they are equivalent!.
Let's do this nicely, would do this as follows
$$
I:=\int x(a^2-x^2)^{1/2}\, dx.
$$
Change of variables 
$$
y=(a^2-x^2)^{1/2},\quad y^2= a^2-x^2,\quad ydy=-xdx.
$$
Note by doing this substitution and squaring y before taking the derivative, we make the integrand something with no fractional exponents, to obtain
$$
I=-\int  y^2 \, dy=-\frac{y^3}{3} +C \to I=-\frac{(a^2-x^2)^{3/2}}{3}+C
$$
where C is a constant of integration.
Now note your method using $ x=a\sin y$ 
$$
I=|a|^3\int  \sin y (1-\sin^2 y)^{1/2} \cos y\, dy=|a|^3\int  \sin y \cos^2 y\, dy.
$$
Change of variables $u=\cos y$, we obtain
$$
I=-|a|^3 \int u^2 \, du = -\frac{|a|^3 u^3}{3}+C=-\frac{|a|^3}{3}\cos^3 y+C=-\frac{|a|^3}{3} \cos^3(\sin^{-1}(\frac{x}{a}))+C=-\frac{|a|^3}{3}(1-x^2)^{3/2}+C.
$$
Your method works, but as you can see is a bit nastier and requires more work.  Hope this helped.
