What is f(x) = int x-cotx dx if f((5pi)/4) = 0 ?

1 Answer
Mar 3, 2017

f(x)=1/2x^2-lnabssinx-(25pi^2+16ln2)/32

Explanation:

f(x)=intx-cotxdx

f(x)=intxdx-intcosx/sinxdx

The first can be integrated using intx^ndx=x^(n+1)/(n+1):

f(x)=1/2x^2-intcosx/sinxdx

For the remaining integral let u=sinx. This implies that du=cosxdx. Then:

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 f((5pi)/4)=0:

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