How to prove this integral in terms of gamma function

Question

Show that $$\Gamma\left(\frac12+x\right) \Gamma\left(\frac12-x\right) = \frac{\pi}{\cos(\pi x)}$$

My approach

$$\Gamma\left(x+\frac12\right) = \frac{(2x!) \sqrt\pi}{2^{2x}x!}$$

I'm stuck here.

Answer 1

In THIS ANSWER, I proved using only real analysis the Euler's Reflection Formula

$$\Gamma(1-y)\Gamma(y)=\frac{\pi}{\sin(\pi y)} \tag 1$$

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Letting $x=y-1/2$ in $(1)$, we see that

$$\Gamma\left(\frac12+x\right)\,\Gamma\left(\frac12-x\right)=\Gamma\left(y\right)\,\Gamma\left(1-y\right)=\frac{\pi}{\sin(\pi y)}=\frac{\pi}{\cos(\pi x)}$$

Answer 2

Euler's reflection formula says that $$\Gamma(z) \cdot \Gamma(1-z)= \frac \pi {\sin(\pi z)}$$

Let $z=\frac 12+x$. Then:

$$\Gamma\left(\frac 12+x\right) \cdot \Gamma \left(\frac 12-x\right)= \dfrac {\pi}{\sin\left(\pi (\frac 12+x)\right)}= \dfrac{\pi}{ \cos(\pi x)}$$