How to prove the inequalities $\int_{0}^{1}\sin{(x^n)}dx\ge\int_{0}^{1}(\sin x)^ndx\ge 0$

Question

Show that: $$\int_{0}^{1}\sin{(x^n)}dx\ge\int_{0}^{1}(\sin x)^ndx\ge 0$$

My idea:maybe $\sin{(x^n)}\ge (\sin{x})^n？$

Answer 1

Let $n\geqslant1$ and $u(x)=\sin(x^n)-(\sin x)^n$ then $$u'(x)=n\cdot\left(x^{n-1}\cos(x^n)-(\sin x)^{n-1}\cos x\right).$$ For every $x$ in $(0,1)$, $x^n\leqslant x$ hence $\cos(x^n)\geqslant\cos x$, and $x\geqslant\sin x$ hence $x^{n-1}\geqslant(\sin x)^{n-1}$. Thus, $u'(x)\geqslant0$ for every $x$ in $(0,1)$. Since $u(0)=0$, this proves that, for every $x$ in $(0,1)$, $$\sin(x^n)\geqslant(\sin x)^n,$$ which yields the comparison of the integrals for $n\geqslant1$ (the fact that both integrals are nonnegative is direct).

Answer 2

Here's an alternative formulation of what is essentially the same as Did's solution, but where the calculus is a little less prominent.

The fact I need for this proof is that $f(u)=\sin u/u$ is positive, less than $1$, and strictly decreasing for $u\in(0,\pi/2)$. That is, for $0<u<v<\pi/2$, we have $1>f(u)>f(v)>0$. And, of course, this can be proven using calculus.

Plug in $u=x^n$, $v=x$ for $x\in(0,1)$, and use that $x^n<x$ and $f(x)^n<f(x)$ since $x,f(x)\in(0,1)$, and we get $$ f(x^n)>f(x)>f(x)^n>0 \quad\text{for all}\quad x\in(0,1). $$ Multiply by $x^n$ and integrate over $(0,1)$ to get $$ \int_0^1\sin{x^n}\,dx=\int_0^1 x^n f(x^n)\,dx >\int_0^1 x^n f(x)^n\,dx=\int_0^1(\sin x)^n >0. $$

Answer 3

Now I have other solution, since $$f(x)=\dfrac{\sin{x}}{x}$$ is Monotone decreasing in $x\in (0,1)$, so $$\dfrac{\sin{x}}{x}\le\dfrac{\sin{x^{n}}}{x^n}$$ so $$x^{n-1}\sin{x}\le \sin{(x^n)}$$ since $$\sin^n{x}=\sin^{n-1}{x}\cdot \sin{x}\le x^{n-1}\sin{x}<\sin{(x^n)}$$