# Integral of $\sin(e^t)dt$

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 $C$.

An integration by parts that gives the first result will also produce the second. Let $u=e^{-t}$, $dv=e^t\sin(e^t)$. Then $du=-e^{-t}dt$, and we can take $v=-\cos(e^t)$. We are left with $$\int_x^{x+1} e^{-t}\cos(e^t)\,dt.$$ Since in our interval we have $e^{-t}\le e^{-x}$, and $|\cos(e^t)|\le 1$, the desired result follows, with $C=1$.
$$\left|\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du\right|=\left|\int_{x}^{x+1} e^{-t} \cos( e^{t})dt\right| \leq \int_{x}^{x+1} e^{-t} dt = -e^t|_{t=x}^{t=x+1} = \left(1-\frac{1}{e}\right)e^{-x}.$$
I think you can see now your constant $C$.