# Prove inequality involving integral

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?

For simplicity, set $$u=t^2$$: $$\int_{0}^{\sqrt{\frac{\pi}{2}}}t^2\cos(t^2)~dt = \dfrac{1}{2}\int_0^{\frac{\pi}{2}}\sqrt{u}cos(u)~du \geq \dfrac{1}{2}\int_{0}^{\sqrt{2}} \sqrt{u}\cos(u)~du \geq \dfrac{1}{2}\int_{0}^{\sqrt{2}} \sqrt{u}\left(1 - \frac{u^2}{2}\right)~du$$
• 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) =(
• 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 $2^{3/4} = (\sqrt{2})^{3/2}$... Feb 9 at 17:21