calculus - How to calculate $int_0^{pi/2} sin^a x cos^b x ,mathrm{d} x$

$$\int_0^{\pi/2} \sin^a x \cos^b x \,\mathrm{d} x = \frac{\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(\frac{1+b}{2}\right)}{2 \Gamma\left(1 + \frac{a + b}{2}\right)} \quad\quad\text{for } a, b > -1$$
according to Mathematica. Wikipedia also lists a recursive expression for the indefinite integral when $a, b > 0$. My question is how to derive the explicit formula given by Mathematica (preferably without using esoteric special functions, but complex analysis is fine).

Answer

Okay first of all we make the substitution $t=\sin^2x$. Thus the integral becomes
$$\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt=\frac12\int_0^1t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}\mathrm dt$$
Then we recall the definition of the Beta function:
$$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\mathrm dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
So we have our integral at
$$\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$