Question

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

Answer 1

`int_0^1xtan^-1xdx ~~ 0.2854` (4dp)

Explanation:

Let `I = int_0^1xtan^-1xdx`

The power series for `tan^-1x` is as follows:

`tan^-1x = x-x^3/3+x^5/5-x^7/7 +x^9/9-x^11/11+x^13/13 - ...`

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

`xtan^-1x = x^2-x^4/3+x^6/5-x^8/7 +x^10/9-x^12/11+x^14/13 - ...`

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

`I ~~int_0^1 x^2-x^4/3+x^6/5-x^8/7 +x^10/9 - ... dx`

Integrating each term gives us:
`I =[ x^3/3-x^5/(3.5)+x^7/(5.7)-x^9/(7.9) +x^11/(9.11) - x^13/(11.13)...]_0^1`

Applying the limits gives us:
`I =1/(1.3)-1/(3.5)+1/(5.7)-1/(7.9) +1/(9.11) - 1/(11.13) + ...`

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 ~~ 0.2854` (4dp)

Answer 2

`int_0^1x(tan)^-1xdx=pi/4-1/2.`

Explanation:

`int_0^1x(tan)^-1xdx`
let`(tan)^-1x=theta`
or `tantheta=x` hence`dx=sec^2theta 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^(pi/4)tan theta sec^2 theta d theta-int_0^(pi/4)(1 inttan theta sec^2 theta d theta)d theta`
=`int_0^(pi/4)tantheta d( tantheta)` -`int_0^(pi/4)((inttantheta d( tantheta)))d theta`
now the indefinite integral `inttantheta d( tantheta)`=`1/2 tan^2 theta`
so for the limit we get the value of first integral=`pi/8`

similarly the indefinite integral `int tan^2 theta d theta`=`int(sec^2 theta -1) d theta`=`tan theta-theta`
so for the req. limit we ge`t =1/2-pi/8`
so the ans. is =`pi/8-(1/2-pi/8)`
=`pi/4-1/2`