Answer:
[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \frac{1}{2}\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)^2(2n)!}} + C[/tex]
General Formulas and Concepts:
Calculus
Integration
- Integrals
- [Indefinite Integrals] Integration Constant C
Sequences
Series
Taylor Polynomials
Power Series
- Power Series of Elementary Functions
- Taylor Series: [tex]\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(c)}{n!}(x - c)^n[/tex]
Integration of Power Series:
- [tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} a_n(x - c)^n[/tex]
- [tex]\displaystyle \int {f(x)} \, dx = \sum^{\infty}_{n = 0} \frac{a_n(x - c)^{n + 1}}{n + 1} + C_1[/tex]
Step-by-step explanation:
*Note:
You could derive the Taylor Series for sin(x) using Taylor polynomials differentiation but usually you have to memorize it.
We are given the integral and are trying to find the infinite series of it:
[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx[/tex]
We know that the power series for sin(x) is:
[tex]\displaystyle sin(x) = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}[/tex]
To find the power series for [tex]\displaystyle \frac{sin(x)}{2x}[/tex], divide the power series by 2x:
[tex]\displaystyle \frac{sin(x)}{2x} = \sum^{\infty}_{n = 0} \bigg[ \frac{(-1)^nx^{2n + 1}}{(2n + 1)!} \cdot \frac{1}{2x} \bigg][/tex]
Simplifying it, we have:
[tex]\displaystyle \frac{sin(x)}{2x} = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{2(2n + 1)!}[/tex]
Rewrite the original integral:
[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \int {\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{2(2n + 1)!}} \, dx[/tex]
Integrate the power series:
[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{2(2n + 1)(2n + 1)!}} + C[/tex]
Simplify the result:
[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \frac{1}{2}\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)^2(2n)!}} + C[/tex]
And we have our final answer.
Topic: AP Calculus BC (Calculus I + II)
Unit: Power Series
