maclaurin series question!? Find the Maclaurin series for $\displaystyle I(x)=\int^{x}_{0}\frac{\sin[t]}{t} \, dt$, and remember to actually show that its a maclaurin series
Any tips/solution on this one? basically I have no idea and all I know about maclaurin series is to expand the terms of the derivatives using the formula: $f^n(0)/n! \cdot x^n$, but idk how to do it in this example
 A: The Maclaurin series is identical to the Taylor series at $x=0$ so since
$$\sin t=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}t^{2n+1}$$
hence integrating term by term gives
$$I(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(2n+1)!}x^{2n+1}$$
A: Sketch: The idea/trick here is to start with the Maclaurin series for $f(t) = \sin t$ and then use what you know about Maclaurin series to compute $I(x)$. Let's say that
$$
\sin(t) = \sum_{n=0}^\infty a_n t^n
$$
(so $a_n = f^{(n)}(0)$). For example, $a_0 = \sin(0) = 0$, and $a_1 = \sin'(0) = \cos(0) = 1$. (Since I'm not sure by your post if you know the general formula, I'll let you find what it is on your own!) Then you can figure out the Maclaurin series for $\frac{\sin(t)}t$ by just distributing the $1/t$ into each term in the series for $\sin(t)$:
$$
\frac{\sin t}t = \frac 1 t \cdot \sin(t) = \frac 1 t \sum_{n=1}^\infty a_n t^n = \sum_{n=1}^\infty a_n t^{n-1}
= \sum_{n=0}^\infty a_{n+1}t^n.
$$
Now to compute $I(x) = \int_0^x \frac {\sin(t)} t \,dt$ we can use a theorem that tells us we may integrate (on the interior of the interval of convergence) the Maclaurin series of $g(x)$ termwise (or term-by-term) to get the Maclaurin series of the $\int_0^x g(t) \,dt$:
$$
    I(x) = \int_0^x \frac{\sin(t)} t \, dt = \int_0^x \left[\sum_{n=0}^\infty a_{n+1}t^{n}\right]\,dt
    = \sum_{n=0}^\infty\left[\int_0^x a_{n+1}t^n\,dt\right].
$$
Note that $n$ doesn't change as $t$ does (that is, $n$ is not a function of $t$), and so when integrating inside each summand we treat $a_{n+1}$ as a constant to find that
$$
I(x) = \sum_{n=0}^\infty\left[\int_0^x a_{n+1}t^n\,dt\right]
     = \sum_{n=0}^\infty\left[\frac{a_{n+1}}{n+1}t^{n+1}\right]\bigg|_{t=0}^x = \sum_{n=0}^\infty \frac{a_{n+1}}{n+1}x^{n+1}.
$$
I wrote this out in general so that you'd determine what $a_n$ is yourself and so that you can see how to do this in the future for other similar problems. Good luck!

[Edit: Revised solution based on OP's comment.]
So that you can do this problem on your own, let me change things to a similar function.
Let's say I want to calculate the Maclaurin expansion of
$$
    I(x) = \int_0^x t^3\cos(t)\,dt.
$$
To do so, I will first compute the Maclaurin series for $f(t) = \cos(t)$. Differentiating I see that
\begin{align*}
    f(0) &= 1\\
    f'(0) &= -\sin(0) = 0\\
    f''(0) &= -\cos(0) = -1\\
    f'''(0) &= \sin(0) = 0\\
    f^{(4)}(0) &= \cos(0) = 1,
\end{align*}
and so on. So the derivatives at $0$ cycle through the values $1,0,-1,0,1,0,-1,0,\ldots$. Writing down only the terms with even index (since the others are $0$) shows that
$$
    \cos t = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n)!}t^{2n}.
$$
Therefore
$$
    t^3\cos(t) = t^3\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n)!}t^{2n}
    = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n)!}t^{2n+3}.
$$
So, since I may integrate term-by-term, I get that
\begin{align*}
    I(x) &= \int_0^x t^3 \cos(t) \,dt\\
&= \sum_{n=0}^\infty \int_0^x\frac{(-1)^{n+1}}{(2n)!}t^{2n+3}\,dt\\
&= \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n)!}\int_0^x t^{2n+3}\,dt\\
&= \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n)!}\cdot \frac{1}{2n+4}x^{2n+4},\\
\end{align*}
and so
$$
I(x) = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n)!(2n+4)}x^{2n+4}.
$$
