Integrate $\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$ I need integrate:
$$\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$$
How can i solve it? is it good way substitute argument of arccos? $$t=\sqrt{\frac{x-4}{x+6}}$$ 

Hint. Here is a route.
By the change of variable 
$$
t=\sqrt{\dfrac{x-4}{x+6}}\qquad x= 2\:\dfrac{ 3 t^2+2}{1-t^2} \qquad dx= 2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'dt
$$ one may obtain
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\int\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'\arccos(t)\:dt
$$ then one may integrate by parts to get
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)\arccos(t)+2\int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt
$$ The latter integral may be evaluated by the change of variable $t:=\sin u$ yielding
$$
\int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt=\frac{10 t}{\sqrt{1-t^2}}-6\: \text{arcsin}(t)+C, \quad |t|<1.
$$
