<s>[INST] 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 
 [/INST] Hint:
$$\dfrac{\cos4x}{\cos3x}=\dfrac{2\cos^22x-1}{\cos x(4\cos^2x-3)}=\dfrac{2(1-2\sin^2x)^2-1}{\cos^2x(4\cos^2x-3)}\cdot\cos x$$
Set $\sin x=u$
$$I=\int_0^{\sqrt3/2}\dfrac{8u^4-8u^2+1}{(1-u^2)(1-4u^2)}du$$
Writing  $u^2=v$ and  use Partial Fraction Decomposition
$$\dfrac{8v^2-8v+1}{(1-v)(1-4v)}=2+\dfrac a{1-v}+\dfrac b{1-4v}$$
Find $a,b$ and replace $v$ with $u^2$
Can you take it home from here?
</s>