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

2 Answers
Nov 2, 2016

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;

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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)

Nov 2, 2016

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 get =1/2-pi/8
so the ans. is =pi/8-(1/2-pi/8)
=pi/4-1/2