Maximize $\int_t^{t+1}\sin(e^x)\,\rm{d}x$ How to maximize the following integral with respect to $t$? 
$$\int_t^{t+1}\sin(e^x)\,\rm{d}x$$ 
My attempt: 
$$\frac{d}{dt}\int_t^{t+1}\sin(e^x)\,\rm{d}x=\sin(e^{t+1})-\sin(e^t)\overset{!}{=}0$$ 
However, I stuck at this step.
 A: $$\sin(e^{t+1})=\sin (e^t)$$
$$e^{t+1}=2n\pi +e^t$$
Use numerical methods to solve for t. 
A: We then have
$$\sin(e^{t+1})=\sin(e^t)$$
$\sin$ is $2\pi$ periodic and hence
$$e^{t+1}=e^t+2\pi n$$
Using exponential rules and letting $T=e^t$ we can then solve this to get
$$eT=T+2\pi n$$
$$(e-1)T=2\pi n$$
$$T=\frac{2\pi n}{e-1}$$
$$t=\ln\frac{2\pi n}{e-1}$$
Alternatively, $\sin(x)=\sin(\pi-x)$, so we have
$$e^{t+1}=\pi(2n+1)-e^t$$
$$eT=\pi(2n+1)-T$$
$$(e+1)T=\pi(2n+1)$$
$$T=\frac{\pi(2n+1)}{e+1}$$
$$t=\ln\frac{\pi(2n+1)}{e+1}$$
Checking a few values, the maximum appears to be $I\simeq0.909026164157$ at $t=\ln(\pi/(e+1))\simeq−0.168531801669$.
For all other $t$, note that by substituting $x\mapsto\ln(x)$ we get
$$I=\int_{e^t}^{e^{t+1}}\frac{\sin(x)}x~\mathrm dx$$
By integrating by parts, this becomes
$$I=\frac{\cos(e^t)}{e^t}-\frac{\cos(e^{t+1})}{e^{t+1}}-\int_{e^t}^{e^{t+1}}\frac{\cos(x)}{x^2}~\mathrm dx$$
This can easily be seen to be bounded and approaching zero very fast, and so checking the first few values of $t$ numerically suffices.
A: I think it doubtful you can do this without numerical methods.  Here is your graph:  
so your maximum is at approximately $t=-0.2$.  
A: $$\sin(e^{t+1})-\sin(e^t)=0$$
$$\frac{(e^{ie^{t+1}}-e^{ie^t})+(e^{-ie^t}-e^{-ie^{t+1}})}{2i}=0$$
if we let $\alpha=e^{ie^t}$ we can get:
$$\alpha(\alpha^{e-1}-1)+\alpha^{-e}(\alpha^{e-1}-1)=0$$
which should be easy to solve numerically.
