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