How to show that $S_{n+1}>S_n$?

Question

Let $S_n = \int_0^{\pi /2}(x\sin{x})^ndx$ for all integer $n\geq1$.

I want to show that $S_{n+1}>S_n$ for all integer $n\geq1$.

Since there is a real $0<a<\frac{\pi}{2}$ such that $a\sin{a}=1$,

we have $\int_a^{\pi /2}(x\sin{x})^{n+1}dx>\int_a^{\pi /2}(x\sin{x})^{n}dx$

but $\int_0^{a}(x\sin{x})^{n+1}dx<\int_0^{a}(x\sin{x})^{n}dx$.

So we can't conclude it is true.

How to prove that $S_{n+1}>S_n$ ?

Thanks in advance.

Answer 1

Let $a$ be the unique real in $(0,\pi/2)$ such that $a\sin a=1$. Now, for $n\ge 1$ we have $$\eqalign{S_{n+1}-S_n&=\int_a^{\pi/2}(x\sin x-1)(x\sin x)^ndx -\int_0^a(1-x\sin x)(x\sin x)^ndx\cr &\ge\int_a^{\pi/2}(x\sin x-1)(x\sin x)dx -\int_0^a(1-x\sin x)(x\sin x)dx\cr &=S_{2}-S_1=\frac{\pi^3}{48}+\frac{\pi}{8}-1\approx0.038>0. }$$ And the desired conclusion follows.

Another approach. Since $$\int_0^{\pi/2}(x\sin x-1)^2(x\sin x)^ndx>0$$ We conclude that $S_{n+2}-2S_{n+1}+S_{n}>0$, so $(S_{n+1}-S_{n})_{n\ge1}$ is increasing, and the conclusion follows from $S_{2}-S_1>0$.