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1

Solved example of power series

$\int\frac{\cos\left(x\right)}{x}dx$
2

Use the Taylor series for rewrite the function $\cos\left(x\right)$ as an approximation: $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$, with $a=0$. Here we will use only the first four terms of the serie

$\int\frac{\frac{1}{0!}+\frac{-1}{2!}x^{2}+\frac{1}{4!}x^{4}+\frac{-1}{6!}x^{6}}{x}dx$
3

Split the fraction $\frac{1+-\frac{1}{2}x^{2}+\frac{1}{24}x^{4}-\frac{1}{720}x^{6}}{x}$ inside the integral, in two terms with common denominator $x$

$\int\left(\frac{1}{x}+\frac{-\frac{1}{2}x^{2}+\frac{1}{24}x^{4}-\frac{1}{720}x^{6}}{x}\right)dx$
4

Split the fraction $\frac{\frac{1}{24}x^{4}-\frac{1}{720}x^{6}}{x}$ in two terms with common denominator $x$

$\int\left(\frac{1}{x}+\frac{-\frac{1}{2}x^{2}}{x}+\frac{\frac{1}{24}x^{4}}{x}+\frac{-\frac{1}{720}x^{6}}{x}\right)dx$
5

Simplifying the fraction by $x$

$\int\left(\frac{1}{x}-\frac{1}{2}x+\frac{1}{24}x^{3}-\frac{1}{720}x^{5}\right)dx$
6

The integral of the sum of two or more functions is equal to the sum of their integrals

$\int\frac{1}{x}dx+\int-\frac{1}{2}xdx+\int\frac{1}{24}x^{3}dx+\int-\frac{1}{720}x^{5}dx$
7

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\int\frac{1}{24}x^{3}dx+\int-\frac{1}{720}x^{5}dx$
8

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\frac{1}{24}\int x^{3}dx+\int-\frac{1}{720}x^{5}dx$
9

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\frac{1}{24}\int x^{3}dx-\frac{1}{720}\int x^{5}dx$
10

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\frac{\frac{1}{24}x^{4}}{4}-\frac{1}{720}\int x^{5}dx$
11

Take $\frac{\frac{1}{24}}{4}$ out of the fraction

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\frac{1}{96}x^{4}-\frac{1}{720}\int x^{5}dx$
12

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\frac{1}{96}x^{4}+\frac{-\frac{1}{720}x^{6}}{6}$
13

Take $\frac{-\frac{1}{720}}{6}$ out of the fraction

$\int\frac{1}{x}dx-\frac{1}{2}\int xdx+\frac{1}{96}x^{4}-\frac{1}{4329}x^{6}$
14

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\int\frac{1}{x}dx-\frac{1}{4}x^2+\frac{1}{96}x^{4}-\frac{1}{4329}x^{6}$
15

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left|x\right|-\frac{1}{4}x^2+\frac{1}{96}x^{4}-\frac{1}{4329}x^{6}$
16

As the integral that we are solving is an indefinite integral, when we finish we must add the constant of integration

$\ln\left|x\right|-\frac{1}{4}x^2+\frac{1}{96}x^{4}-\frac{1}{4329}x^{6}+C_0$

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