Problem Statement
(a) Use the standard expansion for $$(1 + t)^{-1}$$ to show that the Maclaurin series for $$\arctan x$$ is given by $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$$ [4 marks]\n\n(b) Hence, approximate the value of $$\int_{0}^{1/2} \frac{\arctan x}{x} \, dx$$ using the first three non-zero terms. [4 marks]
Verified Solution & Marking Scheme
Part (a): Derivative and Geometric Series
$$\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2} = (1 - (-x^2))^{-1} = 1 - x^2 + x^4 - x^6 + \dots$$ for $$|x| < 1$$.
Integrate term by term
$$\arctan x = \int (1 - x^2 + x^4 - x^6 + \dots)\,dx = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots + C$$. Since $$\arctan(0) = 0$$, $$C = 0$$. (AG)
Part (b): Divide by x
$$\frac{\arctan x}{x} = 1 - \frac{x^2}{3} + \frac{x^4}{5} - \dots$$
Integrate from 0 to 1/2
$$\int_{0}^{1/2} \left(1 - \frac{x^2}{3} + \frac{x^4}{5}\right)\,dx = \left[x - \frac{x^3}{9} + \frac{x^5}{25}\right]_{0}^{1/2} = \frac{1}{2} - \frac{1}{72} + \frac{1}{800} = \frac{3600 - 100 + 9}{7200} = \frac{3509}{7200} \approx 0.487$$