Prove that $|\int_a^b \sin \phi(t) dt| \leq \frac {4}{m}$

Hint:

$$\left|\int_{a}^{b}\dfrac{\sin{(\phi(t))}d(\phi(t))}{\phi'(t)}\right|\le \dfrac{|\cos{\phi{(b)}}-\cos{\phi(a)}|}{m}\le\dfrac{2}{m}$$

$\int_a^b \sin [ \phi (x) ] dx = \int_a^b \{\sin [ \phi (x) ] \phi'(x)\} \frac{1}{ \phi'(x) } dx$

We may apply Theorem 5.5 of Apostol's Calculus Volume 1, with $g(x) = \sin [ \phi (x) ] \phi'(x)$ and $f(x) = \frac{1}{ \phi'(x) }$

Since $\phi''(x)$ is continuous and non-zero it cannot change sign by Bolzano's Theorem (Theorem 3.6 in Apostol's Calculus Volume 1). So $f'(x) = -\frac{\phi''(x)}{\{\phi'(x)\}^2}$ always has the opposite sign as $\phi''(x)$!

The required integral then becomes:

$\frac{1}{\phi'(a)} \int_a^c \sin [ \phi (x) ] \phi'(x) dx + \frac{1}{\phi'(b)} \int_c^b \sin [ \phi (x) ] \phi'(x) dx = \frac{1}{\phi'(a)} \{ \cos(\phi(c)) - \cos(\phi(a)) \} + \frac{1}{\phi'(b)} \{ \cos(\phi(b)) - \cos(\phi(c)) \}$

In view of the fact that $|\cos(x) - \cos(y)| \le 2$ for all values of $x$ and $y$, we have:

$|\int_a^b \sin [ \phi (x) ] dx| \le \frac{4}{m}$.

This completes the proof.