Find the exact value of $t$ that maximizes $\int_{t}^{t+1}\sin e^x \ dx$ The exact value is close to zero judging by the graph. By the fundamental theorem of calculus, $\frac{d}{dt}\int_{t}^{t+1}\sin e^x \ dx=\sin e^{t+1}-\sin e^{t}$. The solution will be given when we solve for $t$ such that $\sin e^{t+1}-\sin e^{t}=0$ . I think we have to play around with the periodicity of sine. So if $\sin e^{t+1}=k$, then $e^{t+1}+2\pi n=k$ and we have $e^t=\frac{k-2\pi n}{e}$ but I can't really work with this.
Other considerations: The inside of the integral can be transformed to $\frac{\sin u}{u}$ by making the substitution $u=e^x$.
Answer: 
judging by the comments I think we an do:
$ee^t=\pi e^t$ So that $e^t=\frac{\pi}{e+1}$ which gives us $t=\ln \frac{\pi}{e+1}$
 A: Let $f(t) = \int_{t}^{t+1}\sin e^x \ dx$. As was noted in the comments, $\sin x - \sin y= 2\sin\left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)$, so $$f'(t) = \frac{d}{dt}\int_{t}^{t+1}\sin e^x \ dx = \sin e^{t+1}-\sin e^t = \\ 2\sin\left(\frac{e^{t+1}-e^t}{2}\right)\cos\left(\frac{e^{t+1}+e^t}{2}\right) = 2\sin\left(e^t\cdot\frac{e-1}{2}\right)\cos\left(e^t\cdot\frac{e+1}{2}\right)$$
So $f'(t) = 0$ when $e^t\cdot\frac{e-1}{2}=n\pi$ or $e^t\cdot\frac{e+1}{2}=n\pi+\pi/2$, or $t = \ln\left(\frac{2\pi n}{e-1}\right)$ or $t = \ln\left(\frac{\pi(2n+1)}{e+1}\right)$ for any $n$.
Now, we want the value closest to zero as you noted, which gives us $\ln(\pi/(e+1))$. $f''(t) = e^t(e\cos e^{t+1} - \cos e^t)$, so $f'' = \frac{\pi}{e+1}\left(e\cos\left(\pi\cdot\frac{e}{e+1}\right)-\cos\left(\frac{\pi}{e+1}\right)\right)$. You can approximate $e$ with $2$ or $3$, but either way $f'' < 0$, so this is a maximum.
There's some more messy algebra if you wish to prove that this is the absolute maximum, if you so desire.
