Show that an integral is more than $\frac{\left(\frac{\pi}{2}\right)^{r+1}}{r+2}$

Question

Q. Consider the function $$f(r)=\int_0^{\frac{\pi}{2}}x^r\sin(x)dx.$$ Show that $f(r)>\frac{\left(\frac{\pi}{2}\right)^{r+1}}{r+2}.$

I do not know how to even begin. Actually, the first part of this question was to show that $f(r)<\frac{\left(\frac{\pi}{2}\right)^{r+1}}{r+1}.$ I was able to do this by noting that for $x\in\left(0,\frac{\pi}{2}\right), x^r\sin(x)<x^r.$ But I do not know how to show that $f(r)>\frac{\left(\frac{\pi}{2}\right)^{r+1}}{r+2}.$

Answer 1

Consider the function $f(x)=x\ cos x -\sin x$. For $x \in (0, \frac{\pi}{2})$, $f'(x)=-x \sin x < 0$. Hence, $f(x)$ is decreasing for $x \in (0, \frac{\pi}{2})$, and as a result, $f(x)< f(0)=0$.

Now, consider the function $g(x)=\frac{\sin x}{x}$. We have:

$$g'(x)=\frac{x \cos x -\sin x}{x^2}=\frac{f(x)}{x^2}< 0;$$

as a result, $g(x)$ is decreasing if $x \in (0, \frac{\pi}{2}).$ Therefore:

$$g(x)=\frac{\sin x}{x} > g(\frac{\pi}{2})=\frac{2}{\pi}. $$

Finally,

$$\frac{2}{\pi}x^{r+1}<x^r \sin x < x^r.$$

The rest is easy.

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PS: There is a post containing the very same inequality at this link.

Answer 2

Hint: Find a simple lower bound for $\dfrac{\sin x}x$ in the interval of interest...