Prove $\int_0^1 x^n \sin x \mathrm{d}x < \frac{1}{n+1}$

Question

Prove $I_n(x)=\int_0^1 x^n \sin x \mathrm{d}x < \frac{1}{n+1}$,

So by now I just found out that by integrating by parts one can get: $$I_n(x)=\cos(1)+n \sin(1)-n(n-1)I_{n-2}(x) \\\leq\\1+n+n(n-1)I_{n-2}(x) \\\leq \\ 1+n-n(n-1)-n(n-1)(n-2)+n(n-1)(n-2)(n-3)I_{n-4}(x) $$ One could iterate this by I cannot really see how gets me closer to the inequality I want to get. On the other hand I have no other ideas how to approach this problem.

Answer 1

For $x \in [0,1]$ we have $ 0 \le x^n \sin x \le x^n$, hence

$\int_0^1 x^n \sin x dx \le \int_0^1 x^n dx= \frac{1}{n+1}.$

It is your turn to show that $\int_0^1 x^n \sin x dx <\frac{1}{n+1}.$

Answer 2

$$\int_0^1x^n\sin x dx<\int_0^1x^n dx=\frac1{n+1}.$$