How do you integrate $int cosxln(sinx)$ using integration by parts?

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Answer 1 · 2016-11-22T20:14:50

It is the same as $int lnt dt$

Let $u = ln(sinx)$ and $dv = cosx dx$ .

This makes $du = 1/sinx cosx dx$ and $v = sinx$

$uv-vdu = sinx ln(sinx) - int sinx(1/sinx cosx) dx$

$= sinx ln(sinx) - int cosx dx$

$= sinx ln(sinx) - sinx +C$ .

Note if you know $int ln t dt$

If you know that $int lnt dt = tlnt - t +C$ , then you can simply substitue $t = sinx$ .

How do you integrate int cosxln(sinx)int cosxln(sinx) using integration by parts?