Taylor Series integraton

**khuezy** · October 8th 2008, 03:33 PM

f(x) = [integral from 0 to x] tan−1(t)dt based at b = 0.

**skeeter** · October 8th 2008, 03:49 PM

$\frac{1}{1+t^2} = \frac{1}{1 - (-t^2)} = 1 - t^2 + t^4 - t^6 + ...$

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 $\arctan{t}$

**khuezy** · October 8th 2008, 03:57 PM

so when there's a problem stating arctan, I immediately change it to 1/(1+x^2)?

**skeeter** · October 8th 2008, 04:09 PM

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

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