Second mean-value theorem for integrals, Apostol 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:
$$\left|\int _a^b \sin(f(x))\,dx\right|\leq \frac{4}{m}$$
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!
 A: $\def\abs#1{\left|#1\right|}$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.
A: A less clever, although valid, proof:
First, observe that
$$
\begin{aligned}
|\cos\alpha-\cos\beta|&=|\cos\alpha+(-\cos\beta)|\\&\le|\cos\alpha|+|-\cos\beta|\\&=|\cos\alpha|+|\cos\beta|\\&\le2.
\end{aligned}
$$
Since $\phi^{\prime\prime}(t)$ is continuous and nonzero on $[a,b]$, then it is strictly positive or negative on $[a,b]$. Now, note that $(1/\phi^{\prime}(t))^{\prime}=-\phi^{\prime\prime}(t)/(\phi^{\prime}(t))^2$, then $(1/\phi^{\prime}(t))^{\prime}$ never changes sign on $[a,b]$. Thus, we can apply Theorem 5.5 for some $\xi\in(a,b)$:
$$
\begin{aligned}
\left|\int_{a}^{b}\sin\phi(t)\,\mathrm{d}t\right|&=\left|\int_{a}^{b}\frac{1}{\phi^{\prime}(t)}\sin(\phi(t))\,\phi^{\prime}(t)\,\mathrm{d}t \right|\\
&=\left|\frac{1}{\phi^{\prime}(a)}\int_{a}^{\xi}\sin(\phi(t))\,\phi^{\prime}(t)\,\mathrm{d}t + \frac{1}{\phi^{\prime}(b)}\int_{\xi}^{b}\sin(\phi(t))\,\phi^{\prime}(t)\,\mathrm{d}t\right|\\
&\le\frac{|\cos\phi(a)-\cos\phi(\xi)|+|\cos\phi(\xi)-\cos\phi(b)|}{m}\\
&\le\frac{4}{m}.
\end{aligned}\\
\hspace{14.5cm}\blacksquare
$$
$
\mbox{}\\
\mbox{}\\
\scriptstyle{\text{Another way to derive the first inequality:}}
$
$$\scriptstyle{ \cos\alpha - \cos\beta = 2\sin\!\left(\frac{\beta+\alpha}{2}\right)\sin\!\left(\frac{\beta-\alpha}{2}\right).} $$
$\scriptstyle{\text{Hence}}$
$$\begin{aligned} \scriptstyle{|\cos\alpha-\cos\beta|} &\scriptstyle{= 2 \left|\sin\!\left(\frac{\beta+\alpha}{2}\right)\right|\left|\sin\!\left(\frac{\beta-\alpha}{2}\right)\right|}\\&\scriptstyle{\le2.} \end{aligned}$$
