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

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- Oct 8th 2008, 03:33 PMkhuezyTaylor Series integraton
f(x) = [integral from 0 to x] tan−1(t)dt based at b = 0.

- Oct 8th 2008, 03:49 PMskeeter
$\displaystyle \frac{1}{1+t^2} = \frac{1}{1 - (-t^2)} = 1 - t^2 + t^4 - t^6 + ...$

integrate term for term ...

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

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

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

o.k. ... integrate $\displaystyle \arctan{t}$ - Oct 8th 2008, 03:57 PMkhuezyok
so when there's a problem stating arctan, I immediately change it to 1/(1+x^2)?

- Oct 8th 2008, 04:09 PMskeeter
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