How to calculate $\int_0^{\pi/2} \sin^a x \cos^b x \,\mathrm{d} x$
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 $$\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$