Question

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

Answer 1

`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`