## Thomas' Calculus 13th Edition

$$\tan^{-1} (\dfrac{2}{3})$$
Recall the Taylor series for $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....; |x| \leq 1$ OR, $=\dfrac{2}{3} - \dfrac{2^3}{3^3 \cdot 3!} + \dfrac{2^5}{3^5 \cdot 5!}-....$ OR, $=\dfrac{2}{3}-( \dfrac{1}{3}) (\dfrac{2}{3})^3+( \dfrac{1}{5}) (\dfrac{2}{3})^5-...$ Or, $=\tan^{-1} (\dfrac{2}{3})$