Beta function problem

Question

Write the following integral in the form of Beta function

$$\int_{0}^{\pi/4} \tan(2x)\, \mathrm{d}x$$

I know that I can use this $$B(p,q)=2 \int_{0}^{\pi/2} \sin^{2p-1}(x) \cos^{2q-1}(x)\, \mathrm{d}x$$

Answer 1

A primitive of $\tan(2x)$ is clearly $-\frac{1}{2}\ln(\cos(2x))$.

This primitive is undefined for $x=-\frac{\pi}{4}$.

Thus the integral is not defined.

Answer 2

Through the formal substitutions $x=\arctan u$ and $u=v^{1/4}$ we get: $$I=\int_{0}^{\pi/4}\tan(2x)\,dx = \int_{0}^{\pi/4}\frac{2\tan(x)}{1-\tan^2(x)}\,dx = \int_{0}^{1}2u(1-u^4)^{-1}\,du \tag{1}$$ from which: $$ I = \frac{1}{2}\int_{0}^{1}v^{-3/4}(1-u)^{-1}\,du = \frac{1}{2}\,B\left(\frac{1}{4},0\right)=\frac{\Gamma\left(\frac{1}{4}\right)\Gamma(0)}{2\,\Gamma\left(\frac{1}{4}\right)}=\frac{\Gamma(0)}{2}\tag{2}$$ but the $\Gamma$ function has a simple pole at $z=0$, hence the integral is diverging. Anyway, I think that to prove the divergence of the integral through that machinery is definitely an overkill.