If $f$ is differentiable and $f'\geq m\geq0$, $|\int_a^b\cos{f(x)}dx|\leq2/m$ 
Suppose $f:[a,b]\to\mathbb R$ is a differentiable function such that its derivative is monotonically decreasing and $f'(x)\geq m>0$ for all $x\in[a,b]$. Prove that $$|\int_a^b\cos f(x)dx|\leq\dfrac{2}{m}$$

I am having some problem with manipulating the modulus and the integral such that I can take $\dfrac{1}{f'(x)}\leq\dfrac{1}{m}$ out. That is, I was thinking about writing it as follows:

Take $f(x)=z$. Then $f'(x)dx=dz$. 

However, I am not quite sure if I can do this because I am not given that $f'$ is also continuous, and change of variable will not work if $f$ is not continuously differentiable. Also, even if I suppose I can do this, I am not quite sure how to proceed. Also, I fail to see why we need the monotonic decreasing property of the derivative.
Some hints will be appreciated.

Since $f'(x) \ge m > 0$, we have $$\int_a^b \cos f(x)\, dx = \int_a^b \frac{1}{f'(x)}f'(x)\cos f(x)\, dx = \int_a^b \frac{1}{f'(x)}\frac{d}{dx}(\sin f(x))\, dx.$$  Since $f'$ is monotonic decreasing and positive, by the Bonnet mean value theorem for integrals, $$\int_a^b \frac{1}{f'(x)} \frac{d}{dx}(\sin f(x))\, dx = \frac{1}{f'(b)}\int_c^b \frac{d}{dx}(\sin f(x))\, dx = \frac{\sin f(b) - \sin f(c)}{f'(b)}$$ for some $c\in [a,b]$. Therefore, $$\left|\int_a^b \cos f(x)\, dx\right| = \frac{|\sin f(b) - \sin f(c)|}{f'(b)} \le \frac{|\sin f(b)| + |\sin f(c)|}{m} \le \frac{2}{m}.$$  
