1
Solved example of power series
$\int\sin\left(x\right)\div xdx$
2
Rewrite the function $\sin\left(x\right)$ as it's representation in Maclaurin series expansion
$\int\frac{\sum_{n=0}^{\infty } x^{\left(2n+1\right)}\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}}{x}dx$
Bring the denominator $x$ inside the power serie
$\int\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}}{x}dx$
Multiplying the fraction by $x^{\left(2n+1\right)}$
$\int\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{x\left(2n+1\right)!}dx$
3
Bring the denominator $x$ inside the power serie
$\int\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{x\left(2n+1\right)!}dx$
Explain more
4
Simplify the fraction $\frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{x\left(2n+1\right)!}$ by $x$
$\int\sum_{n=0}^{\infty } \frac{x^{\left(2n+1-1\right)}{\left(-1\right)}^n}{\left(2n+1\right)!}dx$
We can rewrite the power series as the following
$\sum_{n=0}^{\infty } \frac{1}{\left(2n+1\right)!}\int x^{\left(2n+1-1\right)}{\left(-1\right)}^ndx$
Subtract the values $1$ and $-1$
$\sum_{n=0}^{\infty } \frac{1}{\left(2n+1\right)!}\int x^{2n}{\left(-1\right)}^ndx$
5
We can rewrite the power series as the following
$\sum_{n=0}^{\infty } \frac{1}{\left(2n+1\right)!}\int x^{2n}{\left(-1\right)}^ndx$
Explain more
6
The integral of a function times a constant (${\left(-1\right)}^n$) is equal to the constant times the integral of the function
$\sum_{n=0}^{\infty } \frac{1}{\left(2n+1\right)!}{\left(-1\right)}^n\int x^{2n}dx$
7
Multiplying the fraction by ${\left(-1\right)}^n$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\int x^{2n}dx$
8
Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2n$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\frac{x^{\left(2n+1\right)}}{2n+1}$
9
Multiplying fractions $\frac{{\left(-1\right)}^n}{\left(2n+1\right)!} \times \frac{x^{\left(2n+1\right)}}{2n+1}$
$\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{\left(2n+1\right)!\left(2n+1\right)}$
10
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{\left(2n+1\right)!\left(2n+1\right)}+C_0$
Final Answer
$\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{\left(2n+1\right)!\left(2n+1\right)}+C_0$