Compute the $PV\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$ Problem : 
Evaluate the closed form of : $PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$
Wolfram alpha give me : 
$I=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx=\sqrt{3}-\coth^{-1}{\sqrt{3}}$ 
But i can't get by my try as following : 
$\cos (4x)=8\cos^{4} x-8\cos^{2} x+1$ 
And 
$\cos (3x)=4\cos^{3} x-3\cos x$ 
And I know that 
$PV\displaystyle\int_0^{\frac{π}{3}}\frac{1}{\cos (3x)}dx=0$
So I need to find 
$J=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos^{2} x}{\cos (3x)}dx=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\sin^{2} x}{\cos (3x)}dx$
Now take $y=\cos x$ 
$J=\displaystyle\int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{4x^{3}-3x}dx$ 
From here I don't know how I complete 

Continue with $PV\displaystyle\int_0^{\frac{π}{3}}\frac{dx}{\cos (3x)}=0$ to rewrite the integral as,
$$I =PV \int_0^{\frac{π}{3}}\frac{8(\cos^{2} x-1)\cos^2 x}{4\cos^3 x -3 \cos x}dx
=PV \int_0^{\frac{π}{3}}\frac{-4\sin^{2}x\> d(2\sin x)}{1-4\sin^2 x }dx$$
Substitute $t = 2\sin x$,
$$I =PV \int_0^{{\sqrt3}}\left(1-\frac1{1-t^2}\right)dt
=\sqrt3-\int_0^{1-\epsilon}\frac{dt}{1-t^2}
-\int_{1+\epsilon}^{{\sqrt3}}\frac{dt}{1-t^2}$$
$$=\sqrt3 - \tanh^{-1}t|_0^{1-\epsilon}-\coth^{-1}t|_{1+\epsilon}^{{\sqrt3}}
=\sqrt3 - \coth^{-1}\sqrt3$$
where $(\tanh^{-1}t)'=\frac1{1-t^2}\> \text{for}\> t^2<1$ and $(\coth^{-1}t)'=\frac1{1-t^2}\> \text{for}\> t^2>1$ are used.
