Prove that $\int_{0}^{\frac{\pi}{2}} \sin^{2n+1}(x) \, \ dx \leq \int_{0}^{\frac{\pi}{2}} \sin^{2n}(x) \, \, dx$

Question

I want to show that

$$\int_{0}^{\frac{\pi}{2}} \sin^{2n+1}(x) \, \ dx \leq \int_{0}^{\frac{\pi}{2}} \sin^{2n}(x) \, \, dx$$

where $n \in \mathbb{N}$.

Let $x \in [0,\frac{\pi}{2}]$. Since $0 \leq \sin(x) \leq 1$, we multiply the inequality by $\sin^{2n}(x)$ to get

$$\sin^{2n+1}(x) \leq \sin^{2n}(x)$$

If we take the integral from $0$ to $\frac{\pi}{2}$ on both sides of the inequality, we deduce that

$$\int_{0}^{\frac{\pi}{2}} \sin^{2n+1}(x) \, \ dx \leq \int_{0}^{\frac{\pi}{2}} \sin^{2n}(x) \, \, dx$$

My only concern is the last step. Any insights?

Answer 1

You are doing well!

What you are doing is legit because of a property of the integral:

if $f\le g$ are two maps, then:

$$\int f\leq \int g.$$

Always.

You even have it for strict inequalities.