Prove that $\int_{0}^{\pi /2} \sin(\sin(x))\,\mathrm dx \le 1$
If a function $f$ is Riemann integrable on $[a,b]$, $f$ is bounded on $[a,b]$. If I show $x \mapsto \sin(\sin(x))$ is Riemann integrable then I would know that $f$ is bounded. I can not write $\int \sin(\sin(x)) \mathrm dx$ as a function so, what would I do here?
For $0\le x\le\pi/2 $, you have $0\le\sin x\le x$ (see the geometric definition of sine). Since $\sin(x)$ is monotone in the interval $[0,\pi/2] $ you have the implication $$ 0\le \sin x\le x\le\pi/2 \Rightarrow 0\le \sin(\sin(x))\le\sin (x) $$ hence $$ \int_{0}^{\pi/2}\sin(\sin(x))\; dx\le \int_{0}^{\pi/2} \sin(x)\;dx =1. $$
Because the function has a maximum in $\pi/2$ and because $ x\geq sinx \geq sin(sin(x))$ for $x \in [0,\pi/2]$ then you can use an inequality like this : $$ \int_{0} ^{\pi/2} sin(sin(x)) < sin(1)^2/2+ (\pi/2-sin(1))sin(1) \approx 0.967$$
where the first term on the right represents the area of the triangle from $0$ to $sin(1)$ (given by the function $f(x) = x$ ) and the second one the rectangle from $sin(1)$ to $\pi/2$ with height equal to $sin(1)$ which is the value of the function 's maximum.
