What is #f(x) = int x-cotx dx# if #f((5pi)/4) = 0 #?
1 Answer
Mar 3, 2017
Explanation:
#f(x)=intx-cotxdx#
#f(x)=intxdx-intcosx/sinxdx#
The first can be integrated using
#f(x)=1/2x^2-intcosx/sinxdx#
For the remaining integral let
#f(x)=1/2x^2-int(du)/u#
Which is the natural logarithm integral:
#f(x)=1/2x^2-lnabsu#
#f(x)=1/2x^2-lnabssinx+C#
Use the initial condition
#0=1/2((5pi)/4)^2-lnabssin((5pi)/4)+C#
#0=(25pi^2)/32-ln(1/sqrt2)+C#
Rewriting with log rules:
#0=(25pi^2)/32+1/2ln2+C#
#C=-(25pi^2+16ln2)/32#
So:
#f(x)=1/2x^2-lnabssinx-(25pi^2+16ln2)/32#