TITLE: Prove that: $\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$
QUESTION [6 upvotes]: Prove that: $$\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$$
$0<a<b$
I'm not asking for a complete solution to this, just a hint or a push in the right direction. I've been struggling with this and haven't really made any progress.

REPLY [5 votes]: Integrating by parts, 
$$\int_a^b\frac{\sin x}x\,dx = \frac{\cos a}a-\frac{\cos b}b -\int_a^b\frac{\cos x}{x^2}\,dx.$$
Then apply the triangle inequality (twice) to get
$$\bigg|\int_a^b\frac{\sin x}{x}\,dx\bigg|\leqslant \frac1a+\frac1b+\int_a^b\frac{1}{x^2}\,dx = \frac{2}{a},$$
since $\int_a^b\frac 1{x^2}\,dx = \frac1a-\frac 1b$.