<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] You already have that PV $\int_0^{\pi \over 3}{1 \over \cos 3x} = 0$, so your integral is equal to
$$PV \int_0^{\pi \over 3}{\cos 4x - 1 \over \cos 3x} \,dx$$
Using that ${1 + \cos 4x \over 2} = \cos^2 2x$, this is the 
same as
$$PV \int_0^{\pi \over 3} {2\cos^2 2x - 2\over \cos 3x}\,dx $$
$$= PV \int_0^{\pi \over 3} -{2\sin^2 2x \over \cos 3x}\,dx $$
Now use identities $\cos 3x = 4\cos^3 x - 3\cos x$ and 
$\sin^2 2x = 4\cos^2 x \sin^2 x$. Your integral then becomes
$$PV \int_0^{\pi \over 3} -{8\sin^2 x \cos x \over 4\cos^2 x - 3}\,dx $$
$$=PV \int_0^{\pi \over 3} -{8\sin^2 x \cos x \over 1 - 4\sin^2 x}\,dx $$
Now the subsitution $u = \sin x$ turns this into
$$PV \int_0^{\sqrt{3} \over 2} -{8u^2  \over 1 - 4u^2}\,du $$
$$= PV \int_0^{\sqrt{3} \over 2} 2 - {2 \over 1 - 4u^2}\,du $$
$$= \sqrt{3} - 2PV\int_0^{\sqrt{3} \over 2}{1 \over 1 - 4u^2}\,du$$
$$=\sqrt{3} - PV\int_0^{\sqrt{3} \over 2}{1 \over 1 - 2u} - \int_0^{\sqrt{3} \over 2}{1 \over 1 + 2u}\,du$$
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) - PV\int_0^{\sqrt{3} \over 2}{1 \over 1 - 2u}\,du$$
The principal value integral from $0$ to $1$ here is zero by symmetry of the kernel about $u = {1 \over 2}$, so the above is equal to
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) + \int_{\sqrt{3} \over 2}^1{1 \over 1 - 2u}\,du$$
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) -  \int_{\sqrt{3} \over 2}^1{1 \over 2u - 1}\,du$$
$$= \sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) + {1 \over 2}\ln(\sqrt{3} - 1)$$
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