# Taylor Series integraton

• October 8th 2008, 03:33 PM
khuezy
Taylor Series integraton
f(x) = [integral from 0 to x] tan−1(t)dt based at b = 0.
• October 8th 2008, 03:49 PM
skeeter
$\frac{1}{1+t^2} = \frac{1}{1 - (-t^2)} = 1 - t^2 + t^4 - t^6 + ...$

integrate term for term ...

$\arctan{t} = C + t - \frac{t^3}{3} + \frac{t^5}{5} - \frac{t^7}{7} + ...$

since $\arctan(0) = 0$ ... $C = 0$

$\arctan{t} = t - \frac{t^3}{3} + \frac{t^5}{5} - \frac{t^7}{7} + ...$

o.k. ... integrate $\arctan{t}$
• October 8th 2008, 03:57 PM
khuezy
ok
so when there's a problem stating arctan, I immediately change it to 1/(1+x^2)?
• October 8th 2008, 04:09 PM
skeeter
you're speaking in absolutes, and no one can tell you to always do it this way.

suffice it to say that it's one nice method to find the power series for the arctan function