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

1 Answer
Jim H
Nov 22, 2016

It is the same as #int lnt dt#

Explanation:

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#.

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