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1

Solved example of taylor series

$\int\cos\left(x\right)\cdot\frac{1}{x}dx$
2

Multiplying the fraction and term

$\int\frac{\cos\left(x\right)}{x}dx$
3

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

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

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

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

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

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

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

$\ln\left|x\right|$
6

The integral $\int\frac{1}{x}dx$ results in: $\ln\left|x\right|$

$\ln\left|x\right|$

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

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

Simplifying

$\int-\frac{1}{2}xdx$

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

$-\frac{1}{2}\int xdx$

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

$-\frac{1}{4}x^2$
7

The integral $\int\frac{\frac{-x^{2}}{2}+\frac{1}{24}x^{4}-\frac{1}{720}x^{6}}{x}dx$ results in: $-\frac{1}{4}x^2$

$-\frac{1}{4}x^2$

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

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

Simplifying

$\int\frac{1}{24}x^{3}dx$

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

$\frac{1}{24}\int x^{3}dx$

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

$\frac{1}{24}\frac{x^{4}}{4}$

Simplify the fraction

$\frac{1}{96}x^{4}$
8

The integral $\int\frac{\frac{1}{24}x^{4}-\frac{1}{720}x^{6}}{x}dx$ results in: $\frac{1}{96}x^{4}$

$\frac{1}{96}x^{4}$

Simplify the fraction by $x$

$\int-\frac{1}{720}x^{5}dx$

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

$-\frac{1}{720}\int x^{5}dx$

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

$-\frac{1}{720}\frac{x^{6}}{6}$

Simplify the fraction

$-\frac{1}{4329}x^{6}$
9

The integral $\int\frac{-\frac{1}{720}x^{6}}{x}dx$ results in: $-\frac{1}{4329}x^{6}$

$-\frac{1}{4329}x^{6}$
10

After gathering the results of all integrals, the final answer is

$\ln\left|x\right|-\frac{1}{4}x^2+\frac{1}{96}x^{4}-\frac{1}{4329}x^{6}$
11

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

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

Answer

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

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