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

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

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).
