Question

Find the integral of $\frac{\sin x}{x}$.

Answer 1

Find the integral of the given function

Given function: $\frac{\sin x}{x}$

According to Taylor-Maclaurin's series

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+..............$

$\sin x=\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}$

So,

$\int \frac{\sin x}{x}dx=\int \frac{\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}}{x}dx$

$=\int \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n}}{(2n+1)!}dx$

$=\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!(2n+1)}+c$, where $C$ is the integration constant.

Hence, the integral of $\frac{\sin x}{x}$ is $\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!(2n+1)}+c$ where $C$ is the integration constant.