Evaluate: $\int x\sqrt{\frac{a-x}{a+x}}\hspace{1mm}dx$ 
Evaluate: $$\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x$$

I don't know where to start. Hints/suggestions will be appreciated 
 A: Hint: Let $x=a\sin t$ and $dx=a\cos t dt$
$\displaystyle\int x\sqrt{\dfrac{a-x}{a+x}}dx=\int a\sin t\sqrt{\dfrac{a-a\sin t}{a+a\sin t}} a\cos t dt=a^2\int \sin t\sqrt{\dfrac{1-\sin t}{1+\sin t}} \cos t dt$
Now, multiply the inside of the square root by $\frac{1-\sin t}{1-\sin t}$
$\displaystyle a^2\int \sin t\sqrt{\dfrac{(1-\sin t)^2}{1-\sin^2 t}} \cos t dt= a^2\int \sin t\dfrac{(1-\sin t)}{\cos t}\cos t  dt$
Can you proceed?
A: $$
\frac{a-x}{a+x} = u^2 \rightarrow x = a\frac{1-u^2}{1+u^2} \rightarrow dx = \frac{-4au}{(1+u^2)^2}du
$$
$$
\int x\sqrt{\dfrac{a-x}{a+x}}\hspace{1mm}\mathrm{d}x = \int a\frac{1-u^2}{1+u^2} u\hspace{1mm}\frac{-4au}{(1+u^2)^2}du = -4a^2 \int \frac{u^2(1-u^2)}{(1+u^2)^3}du =
$$
$$
-4a^2 \int \frac{u^2(1-u^2)}{(1+u^2)^3}du = -4a^2[\int \frac{du}{1+u^2} + \int \frac{3du}{(1+u^2)^2} + \int \frac{-2du}{(1+u^2)^3}] 
$$
For each integral apply $u=\tan\theta$
$$
\int \frac{du}{1+u^2} = \arctan(u)
$$
$$
\int \frac{3du}{(1+u^2)^2} = 1/2 (u/(u^2+1)+\arctan(u))
$$
$$
\int \frac{-2du}{(1+u^2)^3} =  1/8 ((u (3 u^2+5))/(u^2+1)^2+3 \arctan(u))
$$
Now substitute $u = \sqrt{\frac{a-x}{a+x}} $
A: This is not the exact problem, but I am sure this will help: http://www.slader.com/textbook/9780618149186-larson-calculus-7th-edition/418/exercises/40b/
Main Idea is to substitute $x = a\sin \theta$ 
