Prove that $|\int_a^b \sin \phi(t) dt| \leq \frac {4}{m}$ If $\phi~''$ is continuous and nonzero on $[a,b]$ and if there is a constant $m>0$ such that $\phi~'(t) \geq m ~\forall~t \in [a,b]$. Prove that $|\int_a^b \sin \phi(t) ~dt| \leq \dfrac {4}{m}$
Attempt: $I = \int_a^b \sin \phi(t)~ dt $
Let $ \phi(t) =u \implies \phi'(t)~dt = du$
Hence, $I = \int_{\phi(a)}^{\phi(b)} \dfrac {\sin u} {u'} du$.
EDIT: I was trying to apply the second mean value theorem for integrals. But, for that, though $\sin$ is continuous in any interval, however, $d(\dfrac {1}{u′})=\dfrac{−u''}{u~′~^2} $, though continuous but may not possess the same sign in the given interval. So, can we still apply the second mean value theorem?
How do I move ahead? Thank you for your help.
 A: Hint:
$$\left|\int_{a}^{b}\dfrac{\sin{(\phi(t))}d(\phi(t))}{\phi'(t)}\right|\le \dfrac{|\cos{\phi{(b)}}-\cos{\phi(a)}|}{m}\le\dfrac{2}{m}$$
A: $\int_a^b \sin [ \phi (x) ] dx = \int_a^b \{\sin [ \phi (x) ] \phi'(x)\} 
\frac{1}{ \phi'(x) } dx$
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.
