Maclaurin serie of $\int_0^x\frac{sin(t)}{t}$ If $f(x)=\int_0^x\frac{\sin(t)}{t}$. Show that
$$f(x)=x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+...$$
Calculate f(1) to three decimal places.
Would you mind showing how to build this Maclaurin serie?
 A: $$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ 
Now, we divide by $x$ to get that 
$$\frac{\sin x}{x} = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$ 
Now, we write
$$f(x)=\int_0^x\frac{\sin tdt}{t}=\int_0^x\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} t^{2n}=\sum^{\infty}_{n=0} \int_o^x\frac{(-1)^n}{(2n+1)!} t^{2n}dt=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} t^{2n+1}\Big|_0^x=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} x^{2n+1}=x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+\cdots$$
To calculate $f(1)$, subsitute $1$ for $x$ into the derived series and hammer through some arithmetic until you find the required accuracy.
A: $$\sin(x) = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots$$
$$\frac{\sin(x)}{x} = 1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots$$
$$\int \frac{\sin(x)}{x}dx = x-\frac{1}{3\cdot 3!}x^3+\frac{1}{5\cdot 5!}x^5-\frac{1}{7\cdot 7!}x^7+\cdots$$
