To determine the power series, start with the geometric series:
#sum_(n=0)^oor^n=1+r+r^2+r^3+...=1/(1-r)#
Let #r=-x^2# and substitute into the series to get:
#sum_(n=0)^oo(-x^2)^n=1+(-x^2)+(-x^2)^2+(-x^2)^3+...=1/(1-(-x^2)#
#->sum_(n=0)^oo(-1)^n x^(2n)=1-x^2+x^4-x^6+...=1/(1+x^2)#
Now integrate this (term by term):
#intsum_(n=0)^oo(-1)^n x^(2n)dx =sum_(n=0)^oo(-1)^n x^(2n+1)/(2n+1) #
#=int1-x^2+x^4-x^6+...dx=x-x^3/3+x^5/5-x^6/6+...#
#=int1/(1+x^2)dx=tan^(-1)(x)#
So, we have determined the series for #tan^-1(x)# to be:
#tan^-1(x)=x-x^3/3+x^5/5-x^7/7+...=sum_(n=0)^oo(-1)^nx^(2n+1)/(2n+1) #
It follows that:
#xtan^-1(x)=x(x-x^3/3+x^5/5-x^7/7+...)#
#=x^2-x^4/3+x^6/5-x^8/7+...=sum_(n=0)^oo(-1)^nx^(2n+2)/(2n+1)#
Therefore:
#int_0^1xtan^-1xdx=int_0^1x^2-x^4/3+x^6/5-x^8/7+...dx=intsum_(n=0)^oo(-1)^nx^(2n+2)/(2n+1)dx#
Integrating term by term gives:
#=[x^3/3-x^5/15+x^7/35-x^9/63+...]_0^1#
Integrating the expression in summation notation gives.
#=[sum_(n=0)^oo(-1)^nx^(2n+3)/((2n+1)(2n+3))]_0^1#
Evaluating the limits, it is clear that this would give:
#=sum_(n=0)^oo(-1)^n/((2n+1)(2n+3))=1/3-1/15+1/35-1/63+...#
Now that we see the pattern in the series we can just keep adding on until we have reached the 4 decimal place condition:
#1/3=0.33333#
#1/3-1/15=0.26667#
#1/3-1/15+1/35=0.29524#
#1/3-1/15+1/35-1/63=0.29524#
#1/3-1/15+1/35-1/63=0.27937#
If we keep going until the value is the same for the 4th decimal place 2 times in a row we get to 52 terms, (note #n=0# is the first term so we wish to sum from #n=0# to #n=51#) so:
#int_0^1xtan^-1xdx~~sum_(n=0)^51(-1)^n/((2n+1)(2n+3))=0.2854#
Analytic evaluation of the integral would yield:
#1/4(pi-2)~~0.285398#
so is in agreement with our power series approximation.
