2
$\begingroup$

Prove that $$\frac{1}{6}<\int_0^1 \frac{1-x^2}{3+\cos(x)}dx < \frac{2}{9}. $$

I tried using known integral inequalities (Cauchy-Schwarz, Chebyshev) but I did not arrive at anything. Then I also tried considering functions of the form $$f(x) = \int_0^x \frac{1-t^2}{3+\cos(t)}dt - \frac{1}{6}$$ and then arrive at something using monotony, but still no answer.

I even tried to compute the integral using the substitution $\displaystyle t = \tan \left(\frac{x}{2} \right),$ but then I arrive at an integral of the form $ \displaystyle \int \frac{(\arctan(x))^2}{x^2+a}dx, a \in \mathbb{R}, $ which I do not know to compute.

$\endgroup$
  • $\begingroup$ Please show your attempt through Cauchy-Schwarz since I am pretty sure it can prove a way sharper inequality than $\frac{3}{18}<I<\frac{4}{18}$. $\endgroup$ – Jack D'Aurizio Oct 7 '18 at 11:06
  • $\begingroup$ @JackD'Aurizio My attempt using Cauchy-Schwarz was merely $\int_0^1 \frac{1-x^2}{3+\cos(x)} dx \leq \sqrt{\int_0^1{(1-x^2)^2}dx \cdot \int_0^1{\frac{1}{(3+\cos(x))^2}}dx}$ but I did not derive the upper bound from this. $\endgroup$ – Cosmin Oct 7 '18 at 11:14
  • 1
    $\begingroup$ An efficient application of Cauchy-Schwarz to $\int_{0}^{1}f(x)\,dx$ comes from writing $f(x)$ as $g(x)\cdot h(x)$ where $g(x)$ and $h(x)$ have an approximately constant ratio on $(0,1)$. It is not the case for $1-x^2$ and $\frac{1}{3+\cos x}$. $\endgroup$ – Jack D'Aurizio Oct 7 '18 at 11:20
2
$\begingroup$

For any $x\in(0,1)$

$$ \frac{1-x^2}{3+\cos x}-\frac{1-x^2}{4} = \frac{(1-x^2)\sin^2\frac{x}{2}}{2(3+\cos x)}\in\left[0,\frac{x^2(1-x)}{12}\right] $$ hence the given integral is bounded between $\frac{1}{6}=\int_{0}^{1}\frac{1-x^2}{4}\,dx$ and $\frac{1}{6}+\frac{1}{144}$.

$\endgroup$
  • $\begingroup$ That's a nice approach! Thank you. $\endgroup$ – Cosmin Oct 7 '18 at 11:21
  • $\begingroup$ I can see $${(1-x^2)\sin^2(x/2)\over2(3+\cos x)}\le{(1-x^2)x^2\over24}$$ (which gives ${1\over6}+{1\over180}$ as an upper bound), but I don't see how you got to $x^2(1-x)/12$. Am I missing something easy and obvious? $\endgroup$ – Barry Cipra Oct 7 '18 at 13:52
1
$\begingroup$

Hint: Since ${\pi\over 3}>1\geq x$ we have $\cos x > {1\over 2}$ so $$3+{1\over 2}<3+\cos x\leq 4$$ so $${1\over 4}\leq {1\over 3+\cos x}< {2\over 7}$$

$${1\over 6}\leq \int_0^1 {1-x^2\over 3+\cos x}dx<{4\over 21}<{2\over 9}$$

$\endgroup$
0
$\begingroup$

This is just a minor variant on greedoid's answer. It suffices to note that $1\lt\pi/2$, so that $0\le\cos x\le1$ for $0\le x\le1$, and thus

$${1\over4}\le{1\over3+\cos x}\le{1\over3}$$

Consequently, since $\int_0^1(1-x^2)dx={2\over3}$, we have

$${1\over6}=\int_0^1{1-x^2\over4}dx\le\int_0^1{1-x^2\over3+\cos x}dx\le\int_0^1{1-x^2\over3}dx={2\over9}$$

The stronger inequality, $1\lt\pi/3$, in greedoid's answer gives, of course, a stronger upper bound (and Jack D'Aurizio's answer gives an even stronger upper bound). But if all you need is $2/9$, the inequality $1\lt\pi/2$ is good enough.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.