Prove that: $\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$ 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.

\begin{align*}
I &= \int_{a}^b \dfrac{\sin x}{x}dx  \\
  &= \dfrac{-\cos x}{x}|_{a}^b+ \int_{a}^b\dfrac{\cos x}{x^2}dx  \\
 &= \dfrac{\cos a}{a} - \dfrac{\cos b}{b}+ J  \\
 &\le \dfrac{1}{a}+\dfrac{1}{b}+  \int_{a}^b \dfrac{1}{x^2}dx  \\
 &= \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{b}  \\
 &= \dfrac{2}{a}  \text{.}
\end{align*}
Also: $I \ge -\dfrac{1}{a} - \dfrac{1}{b}- \displaystyle \int_{a}^b \dfrac{1}{x^2}dx = -\dfrac{2}{a}$. Thus $|I| \le \dfrac{2}{a}$ as claimed.
