General formula for $\int_0^{\pi/2} \tan^{\alpha}(x) dx$? There are already questions about how to find $\int \tan^{1/2}(x) dx$. But how to derive a general formula for $\int_0^{\pi/2} \tan^{\alpha}(x) dx$ (which converges if $|\alpha|<1$) ?
More details: I have to evaluate this integral because I want to derive Euler's reflection formula for gamma functions. The integral above is the result of using a substitution $v=\tan^2(x)$ in $\int_0^\infty \frac{v^\beta}{1+v} dv$ 
 A: 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
$$\int_0^{\frac \pi 2} \tan^\alpha (x)dx=\frac{\pi\sin(\frac {\alpha \pi}2)}{\sin(\alpha \pi)}=\frac{\pi}{2\cos(\frac{\alpha\pi}{2})}$$
A: $$
\begin{align}
\int_0^{\pi/2}\tan^\alpha(x)\,\mathrm{d}x
&=\int_0^\infty\frac{x^\alpha}{1+x^2}\,\mathrm{d}x\tag1\\
&=\frac\pi2\csc\left(\pi\frac{\alpha+1}2\right)\tag2\\[3pt]
&=\frac\pi2\sec\left(\frac{\pi\alpha}2\right)\tag3
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto\arctan(x)$
$(2)$: use this answer
$(3)$: trigonometric identity
A: 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 $|a|<1$. 

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 $n$.
