Integration Proofs

Question

Prove

Show that for $x\geq 1$

$\displaystyle\left|\int_1^x \frac {\sin(t)}{t} dt\right|$ $ \leq \ln(x) $

Attempt:

$\displaystyle\left|\int_1^x \frac {\sin(t)}{t} dt\right|$ $ $ $ \leq $ $\displaystyle\int_1^x$ $\left| \dfrac {\sin(t)}{t}\right| dt$

Answer 1

Note that $\int_{1}^{x}\lvert\frac{\sin(t)}{t}\rvert dt\le\int_{1}^{x}\frac{1}{t}dt$. Now integrate.

Answer 2

You're on the right track. What do you know about the bounds of $\sin(t)$?

Answer 3

In fact, $$ \biggl|\int_{1}^{x}\dfrac{\sin t}{t}dt\biggr| \leq \int_{1}^{x}\biggl|\dfrac{\sin t}{t}\biggr|dt = \int_{1}^{x}\dfrac{|\sin t|}{|t|}dt = \int_{1}^{x}\dfrac{|\sin t|}{t}dt \leq \int_{1}^{x}\dfrac{dt}{t} = \ln x $$ Note that $|\sin x| \leq 1$ for all $x \in \mathbb{R}$.