Proving that $\left|\int_{a}^{b}\frac{\sin(x)}{x}dx\right|\leq 3$, given $1\leq aIf $1\leq a<b$, then
$$ \left|\int_{a}^{b}\frac{\sin(x)}{x}dx\right|\leq 3.$$ 

Proceeding by integration by parts; let $u(x)=\sin(x)$ and $dv(x)=1/x$, then
$u'=\cos(x)$ & $v(x)=\log(x)$. We get, $$\bigg|\sin(x)\log(x)\bigg|_a^b-\int_a^b \cos(x)\log(x)dx\bigg|$$ 
Now, i think it might be convenient to use the fact that $sine$ and $cosine$ are bounded above by 1 and $\log(x)\leq x-1$. Thanks
 A: The stationary points of the sine integral function $\text{Si}(x)$ occur at $x\in\pi\mathbb{Z}\setminus\{0\}$ by the fundamental theorem of calculus, hence the function $\text{Si}(x)$ is bounded between $\text{Si}(1)$ and $\text{Si}(\pi)$ on the interval $(1,+\infty)$. 

It follows that we just have to prove:
$$ \int_{1}^{\pi}\frac{\sin x}{x}\,dx \leq 3 $$
or:
$$ \int_{0}^{\pi-1}\frac{\sin x}{\pi-x}\,dx \leq 3.$$
However, that is trivial, since the LHS is bounded by:
$$ \int_{0}^{\pi-1}\frac{4x}{\pi^2}\,dx = \frac{2(\pi-1)^2}{\pi^2} = \color{red}{0.9294\ldots} $$

A weaker but easier inequality follows from $ \int_{1}^{\pi}\frac{\sin x}{x}\,dx \leq \int_{1}^{\pi} 1\,dx = \pi-1 < 3.$

Your approach also works. By integration by parts:
$$ \text{Si}(b)-\text{Si}(a) = \left.\frac{1-\cos x}{x}\right|_{a}^{b}+\int_{a}^{b}\frac{1-\cos x}{x^2}\,dx $$
but since $0\leq (1-\cos x)\leq 2$,
$$ \left|\text{Si}(b)-\text{Si}(a)\right|\leq \max_{x\geq 1}\frac{1-\cos x}{x}+\int_{1}^{+\infty}\frac{2}{x^2}\,dx < 3. $$
A: Actually, you can improve the result. Noting
$$ \int\frac{\sin x}{x}dx=-\int\frac{1}{x}d\cos x=-\frac{\cos x}{x}+\int\frac{\cos x}{x^2}dx $$
one has
\begin{eqnarray}
\bigg|\int_a^b\frac{\sin x}{x}dx\bigg|&=&\bigg|-\frac{\cos x}{x}|_a^b+\int_a^b\frac{\cos x}{x^2}dx\bigg|\\
&\le&\bigg|-\frac{\cos b}{b}+\frac{\cos a}{a}\bigg|+\int_a^b\frac{1}{x^2}dx\\
&\le&\frac{1}{a}+\frac{1}{b}-\frac{1}{x}\bigg|_a^b\\
&\le&\frac{1}{a}+\frac{1}{b}+\frac{1}{a}-\frac{1}{b}\\
&=&\frac{2}{a}\\
&\le&2.
\end{eqnarray}
