Second mean-value theorem for integrals, Apostol

Question

Let $f''(x)$ be continous and non zero on [a,b] and if $f'(x)\geq m >0,~ \forall x,\in[a,b]$, prove that: $|\int _a^b sin(f(x))dx|\leq \frac{4}{m}$

Now, I need to use this theorem:

$\int_a^bf(x)g(x)dx=f(a)\int_a^cg(x)dx+f(b)\int_c^bg(x)dx$

There is also a hint that I should multiply the integrand by $\frac{f'(x)}{f'(x)}$.

I would be pleased if somebody could help me with directions, of how I should proceed?

Obviously, I should try to leave $f'(x)$ or $\frac{1}{f'(x)}$ in the integrand right? But if I try that I don't seem to get far. Maybe there is an intermediate step? Thanks!

Answer 1

$\def\abs#1{\left|#1\right|}$We have for some $c \in (a,b)$: \begin{align*} \abs{\int_a^b \sin f(x) \, dx} &= \abs{\int_a^b f'(x)\sin f(x) \cdot \frac 1{f'(x)}\, dx}\\ &\le \frac 1{f'(a)}\abs{\int_a^c f'(x)\sin f(x)\, dx} + \frac 1{f'(b)}\abs{\int_c^b f'(x)\sin f(x)\, dx}\\ &\le \frac 1m \abs{\int_{f(a)}^{f(c)} \sin u \, du} + \frac 1m \abs{\int_{f(c)}^{f(b)}\sin u\, du} \end{align*} Now note, that for any interval $[\alpha, \beta]$ we have writing $\beta = \alpha + 2k\pi + \gamma$ where $\gamma < 2\pi$ and $k \in \mathbb N$ \begin{align*} \abs{\int_\alpha^\beta \sin u \, du} &= \abs{\int_0^{\gamma}\sin u \, du}\\ &\le \int_0^{\pi} \abs{\sin u}\, du\\ &= 2. \end{align*} Pluging this into the above gives the desired estimate.