How can I solve approximation evaluation of this integral? 

$$\int_{-1}^{0}\sin(e^{x})\,\text{d}x.$$


Approximation of this formula up to difference (error) $1/5000$.
Because of the error size $1/5000$, I think it's solved by Taylor expansion.
 A: I used the following identity according to the list of Maclaurin series from Wikipedia:
$$\sin(x) = \sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n}{(2n + 1)!}x^{2n + 1} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \quad \text{for all $x$}$$
For your problem, $\sin(e^x)$ can be expressed as
$$\sin(e^x) = \sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n}{(2n + 1)!}(e^x)^{2n + 1} = e^x - \dfrac{e^{3x}}{3!} + \dfrac{e^{5x}}{5!} - \cdots$$
So
$$\begin{aligned}
\int_{-1}^0 \sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n}{(2n + 1)!} e^{x(2n + 1)}\,dx &= \left.\sum\limits_{n = 0}^{\infty}\dfrac{(-1)^n}{(2n + 1)!} \dfrac{e^{x(2n + 1)}}{2n + 1}\right\vert_{x = -1}^{x = 0}
\end{aligned}$$
Equivalently
$$\begin{aligned}
\int_{-1}^0 \sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n}{(2n + 1)!} e^{x(2n + 1)}\,dx &= \int_{-1}^0 \left(e^x - \dfrac{e^{3x}}{3!} + \dfrac{e^{5x}}{5!} - \cdots\right)\,dx\\
&= \left.\left(e^x - \dfrac{e^{3x}}{3 \cdot 3!} + \dfrac{e^{5x}}{5 \cdot 5!} - \cdots \right)\right\vert_{x = -1}^{x = 0}
\end{aligned}$$
By trials, we see that the number of terms in the expression (after the $x$ value is substituted) is 4 terms in order for the approximation difference to be at most $1/5000$.  Hence, the approximation is around $.59804$
A: You are correct : use the classical series expansion of Sin(y), replace y by Exp[x] and integrate between the two bounds. The integrals are very easy to compute since they are all of the form Exp[(2n+1)x].  
You will see that this is very quickly convergent since :    
1 term leads to  0.579331
2 terms lead to  0.580986
3 terms lead to  0.580958   
which is the "exact" value to 6 significant digits.
