Exactly as the title says , we need to prove :

$$\int_\frac{π}{4} ^ \frac{π}{2} e^{\cos x + \cos^2{x}}\mathrm dx>\sqrt2$$

My unsuccessful approach:

I tried by searching for a function $f$ such that :

$$e^{\cos x + \cos^2{x}}>f$$ and

$$\int_\frac{π}{4} ^ \frac{π}{2}f\mathrm dx = \sqrt2$$

Since the limits are in terms of $π$ while the integral should be equal to $\sqrt 2$ , the indefinite integral should be of the form having denominator equal to $π$. Unfortunately, I have never encountered such type of functions or integrals.

For the same reasons above, I also cannot think of functions to apply sandwich theorem.

I am looking for some elementary methods but all level of answers are welcome. Thanks !

  • $\begingroup$ Maybe $e^{\cos(x)+\cos^2(x)}\geq \cos(x)+\cos^2(x)\geq \cos(x)$ ? (I didn't check if the could work... it's just an idea...) $\endgroup$
    – Surb
    May 26 at 18:14
  • $\begingroup$ Where does that problem come from? $\endgroup$
    – Martin R
    May 26 at 18:15
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    $\begingroup$ @Surb Thanks, but $\int_\frac{π}{4} ^ \frac{π}{2} \cos x = 1-\frac{1}{\sqrt {2}}$ and $\int_\frac{π}{4} ^ \frac{π}{2} \cos x + \cos^2x = 3/4 -\frac{1}{\sqrt{2}} + \frac{π}{8}$ which doesn't seem to help :( $\endgroup$ May 26 at 18:19
  • $\begingroup$ @MartinR It was given in my coaching week test. $\endgroup$ May 26 at 18:23
  • $\begingroup$ [NOT AN ANSWER, JUST FOR GRAPHICAL UNDERSTANDNING] desmos.com/calculator/cwlbnucpp2, your integral covers the region above x-axis, between the red strip and below the curve, and using the concept that this area is lesser than the area of the rectangular strip of blue region (basically the blue strip between $\frac{\pi}{4}$ and $\frac{\pi}{2}$ which is 2.618, implies 0.785<$\sqrt 2$<$\int_{{π}/{4}}^{{π}/{2}} e^{\cos x + \cos^2{x}}\mathrm dx$ < 2.618 $\endgroup$
    – ayan
    May 27 at 5:48

3 Answers 3


Let $\,g(x)=\cos x+\cos^2x\,.$

It results that$$e^{g(x)}>1+g(x)+\frac12g^2(x)+\frac16g^3(x)\,.$$ Moreover, $$\int_{\frac{\pi}4}^{\frac{\pi}2}\left[1+g(x)+\frac12g^2(x)+\frac16g^3(x)\right]\!\mathrm dx=\\=\dfrac{7956+3135\pi-6712\sqrt2}{5760}>\sqrt2\;.$$

  • $\begingroup$ How do we know that if I take one more term in the expansion of $e^{g(x)}$, the sum will not go lower? Do we need to prove that? $\endgroup$
    – sku
    May 27 at 1:04
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    $\begingroup$ The last integral is tedious to evaluate. The last inequality is to be observed by doing numerical calculations by hand ? Anyway, thanks for the answer. $\endgroup$ May 27 at 5:02
  • 2
    $\begingroup$ actually $\frac{22}{7}$ > $\pi$ and not $\frac{22}{7}$ < $\pi$, also $\pi$ >3.14 is self - explanatory. refer this for proof en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 (also dont worry much such questions will never appear in contest math) $\endgroup$
    – ayan
    May 27 at 14:22
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    $\begingroup$ @An_Elephant, you are absolutely right when you say that calculating that integral requires more time than $2$ or $3$ minutes. Regarding $\pi$, we know that $\pi=3.14159…$, so it is obvious that $\pi>3.14$, moreover, as User0 pointed out, actually $\frac{22}7>\pi\,.$ $\endgroup$
    – Angelo
    May 27 at 14:36
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    $\begingroup$ @User0 Oops sorry I mis-remembered that integral inequality. Also, it wasn't related to contest math practice. $\endgroup$ May 27 at 14:38

Remarks: The trick is to use $\mathrm{e}^{u} = \mathrm{e}^a\mathrm{e}^{u - a} \ge \mathrm{e}^a (1 + u - a)$ by choosing appropriate constant $a$. More details are given at the end.

We have \begin{align*} I &= \int_{\pi/4}^{\pi/3} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x\\[6pt] &= \int_{\pi/4}^{\pi/3} \mathrm{e}\cdot \mathrm{e}^{\cos x + \cos^2{x} - 1}\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{1/2}\mathrm{e}^{\cos x + \cos^2{x} - 1/2}\,\mathrm{d} x\\[6pt] &\ge \int_{\pi/4}^{\pi/3} \mathrm{e}\cdot (1 + \cos x + \cos^2{x} - 1)\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{1/2}(1 + \cos x + \cos^2{x} - 1/2)\,\mathrm{d} x \tag{1}\\[6pt] &= \mathrm{e}\left(- \frac14 + \frac{1}{24}\pi - \frac{\sqrt 2}{2} + \frac58 \sqrt 3\right) + \mathrm{e}^{1/2}\left(-\frac58\sqrt 3 + \frac16\pi + 1\right)\\[6pt] &> 2.7\left(- \frac14 + \frac{1}{24}\cdot 3.14 - \frac{\sqrt 2}{2} + \frac58 \sqrt 3\right) + 1.64\left(-\frac58\sqrt 3 + \frac16\cdot 3.14 + 1\right)\\[6pt] &> \sqrt 2 \end{align*} where we use $\mathrm{e}^u \ge 1 + u$ for all $u\in \mathbb{R}$ in (1), and $\mathrm{e} > 2.7$, and $\mathrm{e}^{1/2} > 1.64$, and $\pi > 3.14$.

More details for the trick:

First, we try \begin{align*} I &= \int_{\pi/4}^{\pi/2} \mathrm{e}^a\cdot\mathrm{e}^{\cos x + \cos^2{x} - a}\,\mathrm{d} x\\ &\ge \int_{\pi/4}^{\pi/2} \mathrm{e}^a\cdot (1 + \cos x + \cos^2{x} - a)\,\mathrm{d} x\\ &= \mathrm{e}^a \left(\frac34 + \frac38\pi - \frac{\sqrt 2}{2} - \frac{\pi}{4} a\right). \end{align*} Let $$f(a) := \mathrm{e}^a \left(\frac34 + \frac38\pi - \frac{\sqrt 2}{2} - \frac{\pi}{4} a\right).$$ We have $$f'(a) = \frac{\pi}{4}\mathrm{e}^a \left(\frac12 + \frac{3-2\sqrt 2}{\pi} -a \right).$$ Thus, $f(a)$ achieves its global maximum at $a_0 = \frac12 + \frac{3-2\sqrt 2}{\pi}$.

However $f(a_0) < \sqrt 2$. Thus, we split the integral into two parts i.e. $$I = \int_{\pi/4}^{\pi/3} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x$$ and do something similar for each part.

  • $\begingroup$ Thanks ! But I've expected for something more elementary (because our test was basic). Like some nice "$f$" such that $\int f = \sqrt{2}$ directly instead of hard calculations. But thanks for another approach. $\endgroup$ May 27 at 5:07
  • $\begingroup$ @An_Elephant OK. I will try if there is better solution. $\endgroup$
    – River Li
    May 27 at 6:16
  • $\begingroup$ This is a smart and simple (after reading it) solution $\endgroup$ May 27 at 6:50
  • $\begingroup$ @ClaudeLeibovici Thanks. $\endgroup$
    – River Li
    May 27 at 6:52
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    $\begingroup$ @An_Elephant. I am preparing an answer. $\endgroup$ May 27 at 7:10

Since, for the consiered range, the cosine is not negative, using $$e^{\cos(t)+\cos^2(t)}=\sum_{n=0}^\infty a_n\cos^n(t)$$ we shall just add a succession of $\color{red}{\text{positive}}$ terms

The $a_n$ are given by $$a_n=\frac{2 a_{n-2}+a_{n-1}}{n} \qquad \text{with} \qquad a_0=a_1=1$$

On the other side, we know the reduction formula for $$I_n=\int \cos^n(x)\,dx$$ and that $$J_n=\int_{\frac \pi 4}^{\frac \pi 2} \cos^n(x)\,dx ~~> ~0$$

We just need to look at the partial sum $$S_p=\sum_{n=0}^p a_n\, J_n$$

We do not need many terms since $$S_4=\frac{2632-2272 \sqrt{2}+1233 \pi }{2304}=1.42903 ~~>~\sqrt 2$$

  • $\begingroup$ Thanks. But how $S_4$ is equal to the given integral ? Or is it that $S_4<\int_\frac{π}{4} ^ \frac{π}{2} e^{\cos x + \cos^2{x}}\mathrm dx$ ? $\endgroup$ May 27 at 8:34
  • $\begingroup$ @An_Elephant. SInce we only add positive terms, just find $p$ such that $S_p \sqrt 2$. That all. If If I continue the process, we shall be very close to the "exact" value of the integral. $\endgroup$ May 27 at 8:42
  • $\begingroup$ @ClaudeLeibovici The recurrence relation is nice. $\endgroup$
    – River Li
    May 27 at 10:05

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