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The exercise consists of proving the following: $$\frac{4}{21}\cdot 2^{3/4} \leq \int_{0}^{\sqrt{\frac{\pi}{2}}}t^{2}\cos(t^{2})dt$$

With $1-\frac{x^{2}}{2} \leq \cos(x) , \forall x\in [0,\pi ] $ (already proven, got it from Taylor polynomials) my idea was to use something like this $x^{2} (1-\frac{x^{4}}{2}) \leq \int_{0}^{\sqrt{\pi /2}}t^{2}\cos(t^{2})dt$. The problem is I don't see any of either $4/21$ or $2^{3/4}$ showing up.

Any hint or tip on this?

Edit:

I have the solution for this but it seems odd to me. Basically it goes from this

$$cos(x^{2}) \geq 1-\frac{x^{4}}{2}, \ for \ 0 < x < 2^{1/4} \\ cos(x^{2}) \geq 0, \ for \ 2^{1/4} \leq x \leq \sqrt{\pi / 2}$$

and then it proceeds with integration (but from there it's easy). These steps look random to me, plus $cos(x^{2})$ is always greater than $1-\frac{x^{4}}{2}$ so why separate into that two cases?

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For simplicity, set $u=t^2$: $$\int_{0}^{\sqrt{\frac{\pi}{2}}}t^2\cos(t^2)~dt = \dfrac{1}{2}\int_0^{\frac{\pi}{2}}\sqrt{u}cos(u)~du \geq \dfrac{1}{2}\int_{0}^{\sqrt{2}} \sqrt{u}\cos(u)~du \geq \dfrac{1}{2}\int_{0}^{\sqrt{2}} \sqrt{u}\left(1 - \frac{u^2}{2}\right)~du$$

And I'm sure you can take it from here :)

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    $\begingroup$ That change from $\pi / 2 \ to \ \sqrt{2}$ is really badass, wow. I don't see myself coming with this in the middle of the test though (this was an exercise from a test) =( $\endgroup$
    – arpg
    Feb 9 at 17:05
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    $\begingroup$ As you've noticed, at first sight $2^{3/4}$ sounds weird as an outcome from the integral, but not that much when you realize that $u^{3/2}$ will appear after the integration, and $2^{3/4} = (\sqrt{2})^{3/2}$... $\endgroup$
    – FormerMath
    Feb 9 at 17:21
  • $\begingroup$ And what about the solution I posted, do you have anything to say? $\endgroup$
    – arpg
    Feb 9 at 17:34

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