# How do you use a power series to approximate  int_0^1xtan^-1xdx ?

##### 2 Answers
Nov 2, 2016

${\int}_{0}^{1} x {\tan}^{-} 1 x \mathrm{dx} \approx 0.2854$ (4dp)

#### Explanation:

Let $I = {\int}_{0}^{1} x {\tan}^{-} 1 x \mathrm{dx}$

The power series for ${\tan}^{-} 1 x$ is as follows:

${\tan}^{-} 1 x = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + {x}^{9} / 9 - {x}^{11} / 11 + {x}^{13} / 13 - \ldots$

And so we get the integrand by multiplying by $x$ to give:

$x {\tan}^{-} 1 x = {x}^{2} - {x}^{4} / 3 + {x}^{6} / 5 - {x}^{8} / 7 + {x}^{10} / 9 - {x}^{12} / 11 + {x}^{14} / 13 - \ldots$

And so replacing the integrand with the power series we have:

$I \approx {\int}_{0}^{1} {x}^{2} - {x}^{4} / 3 + {x}^{6} / 5 - {x}^{8} / 7 + {x}^{10} / 9 - \ldots \mathrm{dx}$

Integrating each term gives us:
$I = {\left[{x}^{3} / 3 - {x}^{5} / \left(3.5\right) + {x}^{7} / \left(5.7\right) - {x}^{9} / \left(7.9\right) + {x}^{11} / \left(9.11\right) - {x}^{13} / \left(11.13\right) \ldots\right]}_{0}^{1}$

Applying the limits gives us:
$I = \frac{1}{1.3} - \frac{1}{3.5} + \frac{1}{5.7} - \frac{1}{7.9} + \frac{1}{9.11} - \frac{1}{11.13} + \ldots$

We will need to take sufficient terms to get an answer that is stable to 4dp, which I will perform in Excel as follows;

Due to the oscillating natures of the power series it takes a significant number of terms to get a stable answer, but we can conclude that
$I \approx 0.2854$ (4dp)

Nov 2, 2016

${\int}_{0}^{1} x {\left(\tan\right)}^{-} 1 x \mathrm{dx} = \frac{\pi}{4} - \frac{1}{2.}$

#### Explanation:

${\int}_{0}^{1} x {\left(\tan\right)}^{-} 1 x \mathrm{dx}$
let${\left(\tan\right)}^{-} 1 x = \theta$
or $\tan \theta = x$ hence$\mathrm{dx} = {\sec}^{2} \theta d \theta$
then the integral reduces to int_0^(pi/4) theta tantheta sec^2 theta d theta
now we use the integration by parts
so the integral=$\theta {\int}_{0}^{\frac{\pi}{4}} \tan \theta {\sec}^{2} \theta d \theta - {\int}_{0}^{\frac{\pi}{4}} \left(1 \int \tan \theta {\sec}^{2} \theta d \theta\right) d \theta$
=${\int}_{0}^{\frac{\pi}{4}} \tan \theta d \left(\tan \theta\right)$ -${\int}_{0}^{\frac{\pi}{4}} \left(\left(\int \tan \theta d \left(\tan \theta\right)\right)\right) d \theta$
now the indefinite integral $\int \tan \theta d \left(\tan \theta\right)$=$\frac{1}{2} {\tan}^{2} \theta$
so for the limit we get the value of first integral=$\frac{\pi}{8}$

similarly the indefinite integral $\int {\tan}^{2} \theta d \theta$=$\int \left({\sec}^{2} \theta - 1\right) d \theta$=$\tan \theta - \theta$
so for the req. limit we ge$t = \frac{1}{2} - \frac{\pi}{8}$
so the ans. is =$\frac{\pi}{8} - \left(\frac{1}{2} - \frac{\pi}{8}\right)$
=$\frac{\pi}{4} - \frac{1}{2}$