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 

$$\mathrm{PV}\int_0^{\frac \pi3}\frac{\cos 4x}{\cos 3x}\mathrm{d}x=\lim_{\epsilon\to 0^+}\left(\int_0^{\frac \pi 6-\epsilon}\frac{\cos 4x}{\cos 3x}\mathrm{d}x+\int_{\frac \pi 6+\epsilon}^{\frac \pi 3}\frac{\cos 4x}{\cos 3x}\mathrm{d}x\right) $$
and you can find the antiderivative
$$\int \frac{\cos 4x}{\cos 3x}\mathrm{d}x=\int\frac{8\cos^4 x-8\cos^2 x+1}{\cos x(4\cos^2 x-3)}\mathrm{d}x=\int\frac{8\cos^4 x-8\cos^2 x+1}{\cos^2 x(4\cos^2 x-3)}\cos x\,\mathrm{d}x$$
by substituting $u=\sin x$ which transforms the integral to
$$\int\frac{8u^4-8u^2+1}{4u^4-5u^2+1}\mathrm{d}u=\int\left(2+\frac{2u^2-1}{(u-1)(u+1)(2u-1)(2u+1)}\right)\mathrm{d}u $$
and you continue with partial fractions. After you find the antiderivative, plug in the limits of integration, and take limit.
