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
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?
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 $x = a\sin \theta$