Another interesting way to evaluating the integral $\int \frac{\cos(x)}{a+b \cos(x)}\:dx$ I need help in evaluating the following integral, please:
$$\int\frac{\cos(x)}{a+b\cos(x)}dx$$ 
so that we get the following result:
$$
\int \frac{\cos x }{a + b\cos x}\:dx
 = \frac{a}{b\sqrt{a^2-b^2}}
   \arcsin\left(\frac{b+a\cos x}{a+b\cos x}\right)
 - \frac1b
   \arcsin(\cos x)
 + C
$$

Also, I would say I don't have a clue how the substitution can be done. I know another well-known answer which is done using the tangent half-angle substitution, but the result I am asking about exists nowhere online as far as I am concerned. This interesting result is given as a hint to a problem in a physics book and it is completely correct as you can test it yourself. I would appreciate any help. Thanks. 
(edit: I came to know that the substitution [u=cos(x)] would lead to the desired formula. So, maybe you can help with that as a hint.)
 A: Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.
$$I=\int\frac{\cos x}{a+b\cos x}dx$$
$$Ib=\int\frac{b\cos x}{a+b\cos x}dx$$
$$Ib=\int\frac{a+b\cos x}{a+b\cos x}dx-a\int\frac{dx}{a+b\cos x}$$
$$Ib=x-a\int\frac{dx}{a+b\cos x}$$
Then we focus on 
$$J=\int\frac{dx}{a+b\cos x}$$
We may write the integral as
$$J=-\int\frac{\sec^2(\frac x2)}{(b-a)\tan^2(\frac x2)-b-a}dx$$
$$J=\frac1{a+b}\int\frac{\sec^2(\frac x2)dx}{\frac{a-b}{a+b}\tan^2(\frac x2)+1}$$
Then we let 
$$\tan(x/2)=\sqrt{\frac{a+b}{a-b}}u\ \ \Rightarrow\ \  \sec^2(x/2)dx=2\sqrt{\frac{a+b}{a-b}}du$$
Which gives 
$$J=\frac1{a+b}\int\frac{2\sqrt{\frac{a+b}{a-b}}du}{\frac{a-b}{a+b}\big(\sqrt{\frac{a+b}{a-b}}u\big)^2+1}$$
$$J=\frac2{\sqrt{a^2-b^2}}\int\frac{du}{u^2+1}$$
$$J=\frac2{\sqrt{a^2-b^2}}\arctan u$$
$$J=\frac2{\sqrt{a^2-b^2}}\arctan\bigg[\sqrt{\frac{a-b}{a+b}}\tan\bigg(\frac x2\bigg)\bigg]$$
Hence we have 
$$Ib=x-\frac{2a}{\sqrt{a^2-b^2}}\arctan\bigg[\sqrt{\frac{a-b}{a+b}}\tan\bigg(\frac x2\bigg)\bigg]$$
Which means
$$I=\frac{x}b-\frac{2a}{b\sqrt{a^2-b^2}}\arctan\bigg[\sqrt{\frac{a-b}{a+b}}\tan\bigg(\frac x2\bigg)\bigg]+C$$
A: Hint: Use $$u=\tan(\frac{x}{2})$$ so $$du=\frac{1}{2}\sec^2(\frac{x}{2})dx$$ and $$\sin(x)=\frac{2u}{u^2-1},\cos(x)=\frac{1-u^2}{1+u^2}$$ and $$dx=\frac{2du}{1+u^2}$$ and we get
$$\int\frac{2(1-u^2)}{(u^2+1)^2\left(a+\frac{b(1-u^2)}{u^2+1}\right)}du$$
