1
Here, we show you a step-by-step solved example of power series. This solution was automatically generated by our smart calculator:
$\int\left(\frac{\cos\left(x\right)}{x}\right)dx$
2
Rewrite the function $\cos\left(x\right)$ as it's representation in Maclaurin series expansion
$\int\frac{\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}x^{2n}}{x}dx$
3
Bring the denominator $x$ inside the power serie
$\int\sum_{n=0}^{\infty } \frac{\frac{{\left(-1\right)}^n}{\left(2n\right)!}x^{2n}}{x}dx$
Intermediate steps
Multiplying the fraction by $x^{2n}$
$\int\sum_{n=0}^{\infty } \frac{\frac{{\left(-1\right)}^nx^{2n}}{\left(2n\right)!}}{x}dx$
Divide fractions $\frac{\frac{{\left(-1\right)}^nx^{2n}}{\left(2n\right)!}}{x}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{2n}}{x\left(2n\right)!}dx$
Simplify the fraction $\frac{{\left(-1\right)}^nx^{2n}}{x\left(2n\right)!}$ by $x$
$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(2n-1\right)}}{\left(2n\right)!}dx$
4
Simplify the expression
$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(2n-1\right)}}{\left(2n\right)!}dx$
Explain this step further
5
We can rewrite the power series as the following
$\sum_{n=0}^{\infty } \frac{1}{\left(2n\right)!}\int{\left(-1\right)}^nx^{\left(2n-1\right)}dx$
6
The integral of a function times a constant (${\left(-1\right)}^n$) is equal to the constant times the integral of the function
$\sum_{n=0}^{\infty } \frac{1}{\left(2n\right)!}{\left(-1\right)}^n\int x^{\left(2n-1\right)}dx$
Intermediate steps
Multiply the fraction by the term
$\sum_{n=0}^{\infty } \frac{1{\left(-1\right)}^n}{\left(2n\right)!}\int x^{\left(2n-1\right)}dx$
Any expression multiplied by $1$ is equal to itself
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\int x^{\left(2n-1\right)}dx$
7
Multiply the fraction by the term
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\int x^{\left(2n-1\right)}dx$
Explain this step further
Intermediate steps
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 $2n-1$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\frac{x^{\left(2n-1+1\right)}}{2n-1+1}$
Add the values $-1$ and $1$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\frac{x^{\left(2n-1+1\right)}}{2n}$
Add the values $-1$ and $1$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\frac{x^{2n}}{2n}$
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 $2n-1$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\frac{x^{2n}}{2n}$
Explain this step further
9
Multiplying fractions $\frac{{\left(-1\right)}^n}{\left(2n\right)!} \times \frac{x^{2n}}{2n}$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{2n}}{2n\left(2n\right)!}$
10
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{2n}}{2n\left(2n\right)!}+C_0$
Final answer to the problem
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{2n}}{2n\left(2n\right)!}+C_0$