How do you integrate $int x csc^2x$ by integration by parts method?

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Answer 1 · 2018-04-01T08:13:26

The answer is $=-xcotx+ln(|sinx|)+C$

$" Reminder "$

$intcotxdx=ln(|sinx|)+C$

$intuv'=uv-intu'v$

$u=x$ , $=>$ , $u'=1$

$v'=csc^2x$ , $=>$ , $v=-cotx$

Therefore,

$intxsec^2xdx=-xcotx+intcotxdx$

$=-xcotx+ln(|sinx|)+C$

How do you integrate int x csc^2xint x csc^2x by integration by parts method?