Demostrar que $\int_{0}^{\frac{\pi }{2}}\sin(\cos(x))+\cos(\sin(x))dx\leq \frac{\pi ^{2}}{4}$

Question

Que $f:\left [ 0,\frac{\pi }{2} \right ]\rightarrow \mathbb{R}, f(x)=\sin(\cos(x))+\cos(\sin(x))$. Demostrar que %#% $ #%

He conseguido encontrar es menor que $$\int\limits_{0}^{\pi/2}f(x)\,\mathrm{d}x\leq \frac{\pi ^{2}}{4}\,.$ $f$, por lo tanto: $$ \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))$ $ $\left [ 0,\frac{\pi }{2} \right ]$, por lo tanto mi desigualdad no es suficiente para probar la declaración del problema.

Answer 1

Nota $$\int_0^{\pi/2} \cos(\sin(x)) dx = \int_0^{\pi/2}\cos\left(\cos\left(\frac{\pi}{2}-x\right)\right) dx = \int_0^{\pi/2} \cos(\cos(x))dx$ $ tenemos $$\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} $$