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Let $f:\left [ 0,\frac{\pi }{2} \right ]\rightarrow \mathbb{R}, f(x)=\sin(\cos(x))+\cos(\sin(x))$. Prove that $$\int\limits_{0}^{\pi/2}f(x)\,\mathrm{d}x\leq \frac{\pi ^{2}}{4}\,.$$

I've managed to find that $f$ is decreasing on $\left [ 0,\frac{\pi }{2} \right ]$, hence: $$ \int\limits_{0}^{\pi/2} f(x)\,\mathrm{d}x \leq \int\limits_{0}^{\pi/2} f(0)\,\mathrm{d}x = \frac{\pi }{2}\cdot (1+\sin(1))$$ But $1+\sin(1)\geq \frac{\pi }{2}$, thus my inequality isn't enough to prove the problem statement.

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  • $\begingroup$ If $u = \frac{\pi}{2} - x$, the integral becomes $\int_0^{\pi/2} \sin(\sin(u)) + \cos(\cos(u)) \mathrm{d}u$ $\endgroup$
    – wythagoras
    May 4, 2017 at 16:00
  • $\begingroup$ The integral can be expressed in terms of Bessel functions. $\endgroup$
    – Lucian
    Aug 31, 2017 at 21:35

1 Answer 1

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Notice $$\int_0^{\pi/2} \cos(\sin(x)) dx = \int_0^{\pi/2}\cos\left(\cos\left(\frac{\pi}{2}-x\right)\right) dx = \int_0^{\pi/2} \cos(\cos(x))dx$$ We have $$\int_0^{\pi/2}\sin(\cos(x))+\cos(\sin(x))dx = \int_0^{\pi/2}\sin(\cos(x)) + \cos(\cos(x)) dx\\ = \int_0^{\pi/2}\sqrt{2}\sin\left(\cos(x)+\frac{\pi}{4}\right) dx \le \int_0^{\pi/2}\sqrt{2} dx = \frac{\pi}{2}\sqrt{2} \le \frac{\pi}{2}\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} $$

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