General formula for $\int_0^{\pi/2} \tan^{\alpha}(x) dx$?

With $u=\tan x$, we have $$\int_0^{\frac \pi 2} \tan^\alpha (x)dx=\int_0^{+\infty}\frac{u^\alpha}{1+u^2}du$$ Now, according to this question: $$\int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\beta +x^{2}}=\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }$$ So, plugging in $\beta=\frac \pi 2$ yields $$\int_0^{\frac \pi 2} \tan^\alpha (x)dx=\frac{\pi\sin(\frac {\alpha \pi}2)}{\sin(\alpha \pi)}=\frac{\pi}{2\cos(\frac{\alpha\pi}{2})}$$

$$ \begin{align} \int_0^{\pi/2}\tan^\alpha(x)\,\mathrm{d}x &=\int_0^\infty\frac{x^\alpha}{1+x^2}\,\mathrm{d}x\tag1\\ &=\frac\pi2\csc\left(\pi\frac{\alpha+1}2\right)\tag2\\[3pt] &=\frac\pi2\sec\left(\frac{\pi\alpha}2\right)\tag3 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto\arctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity

Consider the integral $$I(a,b)=\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx,\qquad a,b>-1$$ Setting $t=\sin(x)^2$: $$I(a,b)=\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt$$ Then recall the definition of the beta function $$\mathrm{B}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ We use this to see that $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ Hence $$\int_0^{\pi/2}\tan(x)^a\mathrm dx=\frac12\Gamma\left(\frac{1+a}2\right)\Gamma\left(\frac{1-a}2\right)$$ Then using $$\Gamma(1-s)\Gamma(s)=\frac{\pi}{\sin\pi s}$$ It can be easily shown that $$\Gamma\left(\frac{1+s}2\right)\Gamma\left(\frac{1-s}2\right)=\pi\sec\frac{\pi s}2$$ So $$\int_0^{\pi/2}\tan(x)^a\mathrm dx=\frac\pi2\sec\frac{\pi a}2$$ Which you know works for $|a|<1$.

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This can be used to show that $$\int_0^{\pi/2} \log^{n}[\tan x]\mathrm dx=\frac\pi2\left(\frac{d}{da}\right)^n\sec\frac{\pi a}2\,\bigg|_{a=0}$$ Or simply $$\int_0^{\pi/2} \tan(x)^{a}\log^{n}[\tan x]\mathrm dx=\frac\pi2\left(\frac{d}{da}\right)^n\sec\frac{\pi a}2$$ To show this just take $\left(\frac{d}{da}\right)^n$ on both sides for integer $n$.