How do I prove this trigonometric integral inequality? If f is integrable and monotone on [a,b] then 
$\left |\int^b_a f(x)\cos x\,dx\right | \le 2(|f(a)-f(b)|+|f(b)|).$
I've tried integration by parts and using the integral inequality property but I'm not sure how to piece everything together. I figure that I need to somehow use the fact that $|\sin x|\, \le 1$ but I'm not sure how. Any help would be appreaciated. Thanks.
 A: We have 
\begin{align}\left|\int_a^b f(x) \cos x\, dx \right|&= \left|\int_a^b [f(x) - f(b)]\cos x\, dx + \int_a^b f(b)\cos x\, dx\right|\\
&\le \left|\int_a^b [f(x) - f(b)]\cos x\, dx\right| + \left|\int_a^b f(b)\cos x\, dx\right|\\
&= |f(c) - f(b)|\left|\int_a^b \cos x\, dx\right| + |f(b)|\left|\int_a^b \cos x\, dx\right|\quad (\text{some $c\in (a,b)$})\\
&= (|f(c) - f(b)| + |f(b)|)\left|\int_a^b \cos x\, dx\right|\\
&= (|f(c) - f(b)| + |f(b)|)|\sin b - \sin a|\\
&\le 2(|f(a) - f(b)| + |f(b)|) \quad (\text{by montonicity of $f$}).
\end{align}
A: It looks like you are only a small step away from solving the problem. Here is how I would do: First integrate by parts:
$$\left| \int_a^b f(x) \cos x \,dx \right|
=\left| f(b)\sin b -f(a) \sin (a) - \int_a^b f'(x) \sin \,x dx  \right| .$$
Then use the triangle inequality and additionally
$$\left| \int_a^b f'(x) \sin \,x dx  \right|
\leq \int_a^b |f'(x)| |\sin \,x| dx \leq \int_a^b |f'(x)|dx = \left| \int_a^b f'(x)\,dx\right|,$$
where the last step is true because $f$ is monotone.
I guess you can piece the rest together.
To get the inequality on the exact form in your question, you can use the fact that
$$f(b) \sin(b) -f(a) \sin(a) = f(b) (\sin b -\sin a) + [f(b)-f(a)] \sin a.$$
