$\int_{0}^{\pi/2} (\sin x)^{1+\sqrt2} dx$ and $\int_{0}^{\pi/2} (\sin x)^{\sqrt2\space-1} dx $

Question

How do I evaluate $$\int_{0}^{\pi/2} (\sin x)^{1+\sqrt2} dx\quad \text{ and }\quad \int_{0}^{\pi/2} (\sin x)^{\sqrt2\space-1} dx \quad ?$$

Answer 1

$$\beta(x,y) = 2 \int_0^{\pi/2} \sin^{2x-1}(a) \cos^{2y-1}(a) da \implies \int_0^{\pi/2} \sin^{m}(a) da = \dfrac{\beta((m+1)/2,1/2)}2$$ Hence, $$\int_0^{\pi/2} \sin^{1+\sqrt2}(a) da = \dfrac{\beta(1+1/\sqrt2,1/2)}2$$ $$\int_0^{\pi/2} \sin^{\sqrt2-1}(a) da = \dfrac{\beta(1/\sqrt2,1/2)}2$$