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.

  • $\begingroup$ 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 '17 at 16:00
  • $\begingroup$ The integral can be expressed in terms of Bessel functions. $\endgroup$ – Lucian Aug 31 '17 at 21:35

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


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.