Integrate $\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$

Question

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? $$t=\sqrt{\frac{x-4}{x+6}}$$

Answer 1

Here is another route:

First integrating by parts, writing the integral as $$ \int 1\cdot \arccos\sqrt{\frac{x-4}{x+6}}\,dx $$ will leave you with an integral like (I forget about constants, and suggest you to do the calculations!) $$ \int \frac{x}{\sqrt{x-4}(x+6)}\,dx. $$ Adding and subtracting $6$ in the numerator (and integrating the part looking like $1/\sqrt{x-4}$) will leave you with something like $$ \int\frac{1}{\sqrt{x-4}(x+6)}\,dx $$ Now, let $t=\sqrt{x-4}$ and you will get something like (again, check constants) $$ \int\frac{1}{t^2+10}\,dt, $$ which will just give an arctan.

If you want to know what I actually got in the end, move your mouse over the box below.

$$ x \arccos\sqrt{\frac{x-4}{x+6}}+\sqrt{10(x-4)}-6\arctan\sqrt{\frac{x-4}{10}}+C$$

Answer 2

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 $$ \int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt=\frac{10 t}{\sqrt{1-t^2}}-6\: \text{arcsin}(t)+C, \quad |t|<1. $$