Ratio of Trigonometric Integrals

$\boxed{1.-}$

$n \ge 1$ , $\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2n + 1} (x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin^{2n}(x) \cdot \sin (x)\, dx =$ Integrating by parts, $= \left(- \cos(x) \cdot \sin^{2n}(x) \right)_{0}^{\pi/2} + 2n \cdot \int_{0}^{\frac{\pi}{2}} \sin^{2n - 1}(x) \cdot \cos^{2}(x)\, dx =$ ( $\cos^2(x) = 1 - \sin^2(x)$ ) $2n \cdot (\int_{0}^{\frac{\pi}{2}} \sin^{2n - 1}(x) \, dx - \int_{0}^{\frac{\pi}{2}} \sin^{2n + 1}(x) \, dx) \Rightarrow$ Let's call $\displaystyle I = \int_{0}^{\frac{\pi}{2}} \sin^{2n + 1} (x) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} \sin^{2n + 1} (x) \, dx = \frac{2n}{2n +1} \cdot \int_{0}^{\frac{\pi}{2}} \sin^{2n - 1}(x) \, dx \Rightarrow$ $\displaystyle \frac{\int_{0}^{\frac{\pi}{2}} \sin^{2n + 2} (x) \, dx}{\int_{0}^{\frac{\pi}{2}} \sin^{2n} (x) \, dx} = \frac{2n +1}{2n +2} =\frac{2n + 2 - 1}{2n +2} = 1 - \frac{1}{2n + 2}$

$\boxed{2.-}$

You can also use Beta function as suggested by @Aditya Sharma

$\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2n + 2} (x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin^{2(n + 3/2) - 1} (x) \cdot \cos^{2(1/2) - 1} (x)\, dx =$ $\frac{B(n + 3/2, 1/2)}{2} = \frac{\Gamma(n + 3/2) \cdot \Gamma (1/2)}{2 \cdot \Gamma(n + 2)}$

In the same way, $\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2n} (x) \, dx = \frac{B(n + 1/2, 1/2)}{2} = \frac{\Gamma(n + 1/2) \cdot \Gamma (1/2)}{2 \cdot \Gamma(n + 1)}$

This implies that, $\displaystyle \frac{\int_{0}^{\frac{\pi}{2}} \sin^{2n + 2} (x) \, dx}{\int_{0}^{\frac{\pi}{2}} \sin^{2n} (x) \, dx} = \frac{\frac{\Gamma(n + 3/2) \cdot \Gamma (1/2)}{2 \cdot \Gamma(n + 2)}}{\frac{\Gamma(n + 1/2) \cdot \Gamma (1/2)}{2 \cdot \Gamma(n + 1)}} = \frac{(n + 1/2)}{(n + 1)} = \frac{(n + 1 - 1/2)}{(n + 1)} = 1 - \frac{1}{2n + 2}$

A quicker way would be to put n=2 and verify the options