How do you integrate #(tanx)ln(cosx)dx#?

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How do you integrate #(tanx)ln(cosx)dx#?

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Answer 1 · 2015-09-13T20:56:46

#-ln^2(cosx)/2+C#

Explanation:

Lets choose: #u=ln(cosx)#, then:
#du=(ln(cosx))'dx=(1/cosx)(cosx)'dx=(1/cosx)(-sinx)dx=-sinx/cosxdx=-tanxdx#
So, #tanxdx=-du# and
#inttanxln(cosx)dx=int(-du)u=-u^2/2+C=-ln^2(cosx)/2+C#

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How do you integrate #(tanx)ln(cosx)dx#?

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