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 
 A: $$PV\ \ 2\int_{-\pi/3}^{\pi /3}\frac{\cos (4x)}{\cos (3x)}dx= \lim_{y\to 0^+}\int_{-\pi/3}^{\pi /3}\frac{\cos (4(x+iy))}{\cos (3(x+iy))}dx+\int_{-\pi/3}^{\pi /3}\frac{\cos (4(x-iy))}{\cos (3(x-iy))}dx$$
Then expand  $\frac1{1+e^{6(x\pm iy)}}$ in geometric series and invert $\int,\sum$, you'll obtain a series of the form $\sum_m \frac{a_m}{m}$ with $a_m$ periodic, which is a finite sum of $\log$
A: 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?
A: $$\mathrm{PV}\int_0^{\frac \pi3}\frac{\cos 4x}{\cos 3x}\mathrm{d}x=\lim_{\epsilon\to 0^+}\left(\int_0^{\frac \pi 6-\epsilon}\frac{\cos 4x}{\cos 3x}\mathrm{d}x+\int_{\frac \pi 6+\epsilon}^{\frac \pi 3}\frac{\cos 4x}{\cos 3x}\mathrm{d}x\right) $$
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.
A: 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.
A: 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
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) + \int_{\sqrt{3} \over 2}^1{1 \over 1 - 2u}\,du$$
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) -  \int_{\sqrt{3} \over 2}^1{1 \over 2u - 1}\,du$$
$$= \sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) + {1 \over 2}\ln(\sqrt{3} - 1)$$
