Baby rudin chapter 6 exercise 14 ---Isn't it a typo? $$
f(x) \ = \ \int_{x}^{x+1} \sin(e^t) \, dt.
$$
Show that 
$$
e^x | f(x) | \ &lt; \ 2
$$
and that 
$$
e^x f(x) = \cos(e^x) - e^{-1} \cos(e^{x+1}) + r(x), 
$$
where $|r(x)|&lt; C e^{-x}$ for some constant $C$.

Using Integration by parts, I showed that
$$
f(x) = \frac{\cos(e^x)}{e^x} - \frac{\cos(e^{x+1})}{e^{x+1}} - \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} \, du .
$$
Thus, in my calculation, 
$$
r(x) \ = \ -e^x \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} \, du .
$$
But I can't prove that $|r(x)| &lt; C r^{-x}$ for some $C$.

Is it typo???? 
(We can show that $|r(x)| &lt; 2 e^{-1}$.)
let $e^x=u$,then
\begin{align*}
|f(x)|&amp;=|\int_{x}^{x+1}\sin{(e^t)}dt|=|\int_{e^{x}}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|\\
&amp;=\left|\dfrac{\cos{e^{x+1}}}{e^{x+1}}-\dfrac{\cos{e^x}}{e^x}+\int_{e^{x}}^{e^{x+1}}\dfrac{\cos{u}}{u^2}du\right|\\
&amp;\le|\dfrac{\cos{e^{x+1}}}{e^{x+1}}|+|\dfrac{\cos{e^x}}{e^x}|+\int_{e^x}^{e^{x+1}}\dfrac{1}{u^2}du\\
&amp;=\dfrac{2}{e^x}
\end{align*}
let $e^x=u$,then
\begin{align*}
|f(x)|&amp;=|\int_{x}^{x+1}\sin{(e^t)}dt|=|\int_{e^{x}}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|\\
&amp;=\left|\dfrac{\cos{e^{x+1}}}{e^{x+1}}-\dfrac{\cos{e^x}}{e^x}+\int_{e^{x}}^{e^{x+1}}\dfrac{\cos{u}}{u^2}du\right|\\
&amp;\le|\dfrac{\cos{e^{x+1}}}{e^{x+1}}|+|\dfrac{\cos{e^x}}{e^x}|+\int_{e^x}^{e^{x+1}}\dfrac{1}{u^2}du\\
&amp;=\dfrac{2}{e^x}
\end{align*}