I am having a little trouble understanding this part of a proof.

There is an integral $\text{J}_{n} = \int_0^{\frac{\pi}{2}} x^2\cos^{2n}x dx$

Now, $\text{J}_0 = \frac{\pi ^3}{24} $

The part of the proof I am unable to understand says that $x \lt \frac{\pi}{2}\sin x $ for every $0 \lt x \lt \frac{\pi}{2}$

They are saying that thus inequality implies that $$\text{J}_n \lt \frac{\pi^2}{4} \int_0^{\frac{\pi}{2}}\sin^2x \cos^{2n} x dx $$

I am unable to supply the intermediate steps. Please help.

  • $\begingroup$ That seems a typo. The correct inequality you would like to apply is $x < \frac{\pi}{2}\sin x$ when $0<x<\frac{\pi}{2}$. $\endgroup$ – Sangchul Lee Mar 13 '16 at 5:42
  • $\begingroup$ @SangchulLee you are right. It was a typo. I corrected it but I still don't know how to fill in the steps. $\endgroup$ – user230452 Mar 13 '16 at 5:47
  • $\begingroup$ Did you get the part where this inequality implies your last inequality involving $\mathrm{J}_n$? It is straightforward, so I assume you did and you are asking about the inequality $x < \frac{\pi}{2}\sin x$. This follows from the strict convexity of the sine function on the interval $[0, \pi/2]$. $\endgroup$ – Sangchul Lee Mar 13 '16 at 5:48
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    $\begingroup$ If I am not misunderstood, you can just bound your integrand using $$ x^2 \cos^{2n} x < (\tfrac{\pi}{2}\sin x)^2 \cos^{2n}x, \quad 0< x < \tfrac{\pi}{2}. $$ $\endgroup$ – Sangchul Lee Mar 13 '16 at 6:16
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    $\begingroup$ Sure. If in addition they are non-negative functions, you may as well recognize it as an inequality between areas under the graph of $A$ and $B$. That is, it is simply read as 'larger shape has bigger area'. $\endgroup$ – Sangchul Lee Mar 13 '16 at 6:23

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