Find the value of the integral $\int_0^1 \frac{\sin(x)}{x}dx$ Given the integral
$$\int_0^1 \frac{\sin(x)}{x}dx$$
I have to find the value of the integral with an accuracy better than $10^{-4}$, however, I am pretty lost.
I am given a hint saying that I should replace $\sin(x)$ with the Taylor polynomium of order $2n$ and evaluate the remainder $Rnf(x)$. I know that the Taylor series is given by
$$
\sum_{n=1}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n
$$
so by replacing with $2n$ we have
$$
\sum_{n=1}^\infty \frac{f^{(2n)}(x_0)}{2n!}(x-x_0)^{2n}
$$
but am I really expected to calculate this? And if so how would I even do this?
Can you help me in the right direction? Thanks in advance.
 A: We know that the Taylor Series for $\sin x$ is
$$
\sin x = \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}
$$
Then the Taylor Series for $\dfrac{\sin x}{x}$ is 
$$
\dfrac{\sum_{n=0}^\infty (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}}{x} = \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n}}{(2n+1)!}
$$
Then we have
$$
\begin{aligned}
\int_0^1 \dfrac{\sin x}{x} \;dx&= \int_0^1 \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n}}{(2n+1)!} \;dx \\
&=\sum_{n=0}^\infty \int_0^1 (-1)^n \dfrac{x^{2n}}{(2n+1)!} \;dx \\
&= \sum_{n=0}^\infty (-1)^n \dfrac{x^{2n+1}}{(2n+1)(2n+1)!} \bigg|_0^1 \\
&= \sum_{n=0}^\infty \dfrac{(-1)^{n}}{(2n+1)(2n+1)!}
\end{aligned}
$$
But this is an Alternating Series so the error in summing the first $k$ terms (compared to the final infinity sum) is at most the magnitude of the $k+1$ term. You want the error to be at most $10^{-4}=0.0001$. So you just want a $k$ so that
$$
\bigg|\dfrac{(-1)^{n}}{(2k+1)(2k+1)!}\bigg| < 0.0001
$$
Trying a few terms we find that $k=3$ works. So we need only sum
$$
\sum_{n=0}^{2} \dfrac{(-1)^{n}}{(2n+1)(2n+1)!}= 1-\dfrac{1}{18}+\dfrac{1}{600} \approx 0.946111
$$
We can check this as
$$
\int_0^1 \dfrac{\sin x}{x} \;dx \approx 0.9460830703671841
$$
Note that I used the properties of Alternating Series, but here this is equivalent to using Taylor's Remainder Theorem, that if you only sum the first $k$ terms of a convergent Taylor Series, the magnitude error in the sum is at most
$$
M \dfrac{R^{k+1}}{(k+1)!}
$$
where $R$ is the distance from the center you chose for the Taylor Series and the point you are evaluating at and $M$ is a bound on your derivative, i.e. if you are using the Taylor Series centered at $x=a$ and you are evaluating the series at $x=b$, then $R= |b-a|$ and $M$ is a number so that $|f^{(k+1)}(x)| \leq M$ for all $x$ between $a$ and $b$. 
A: On taking $x_0=0$, we get the taylor polynomial of $\sin x$ to be
$$\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots+\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}+R_n(x)=\sum_{k=0}^n\dfrac{(-1)^kx^{2k+1}}{(2k+1)!}+R_n(x)$$
Dividing it by $x$ and integrating we get
$$\int_0^1\dfrac{\sin x}xdx = \int_0^1\left(\sum_{k=0}^n\dfrac{(-1)^kx^{2k}}{(2k+1)!}+R_n(x)\right)dx\\
=\sum_{k=0}^n\dfrac{(-1)^k}{(2k+1)(2k+1)!}+\int_0^1R_n(x)dx$$
Now note that $R_n(x)=O(x^{2n+3})$ and therefore, we can safely ignore this term after a good approximation is reached.
A: I am not sure if this is allowed but if someone in the future might need help to understand these particular problems, in general, the following Youtube-video made it so much clearer:
https://www.youtube.com/watch?v=3ZOS69YTxQ8
