Let $f(x) = \int_x^{x+1} \sin(e^t)dt$. Show that $ e^x|f(x)| < 2 $ and that $ e^xf(x) = \cos(e^x) - e^{-1}\cos(e^{x+1}) + r(x) $ where $|r(x)| < Ce^{-x}$ for some constant $C$.
I can do the first part easily. Integration by parts gives us $ 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, $ so substituting $\cos u = 1$ and $\cos u = -1$ gives \begin{align*} \cos e^x - 1 + \frac{1}{e}\left(1- \cos e^{x+1}\right) &\leq e^xf(x) \leq \cos e^x + 1 - \frac{1}{e}\left(\cos e^{x+1} + 1\right)\\ -2 &< e^xf(x) < 2 \end{align*} So that $e^x|f(x)| < 2$.
I get stuck on the second part, though. From integration by parts, I have $e^xf(x) = \cos e^x - e^{-1}\cos e^{x+1} - e^x\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du$, so it reduces to showing that $ \left|\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du\right| < Ce^{-2x} $ for some constant $C$. However, I always end up with a constant for $Ce^{x}$ rather than for $C$.