`int tanxln(cosx) dx` Find the indefinite integral

`int tanx ln(cosx)dx`

To solve, apply u-substitution method.

`u=ln(cosx)`

`du=1/(cosx)*(-sinx)dx`

`du =-tanxdx`

`-du=tanxdx`

Expressing the integral in terms of u variable, it becomes

`= int ln(cosx) * tanxdx`

`= int u * (-du)`

`=-int udu`

To take the integral of this, apply the formula `int x^n dx =x^(n+1)/(n+1)+C` .

`= -u^2/2 + C`

And, substitute back u = ln(cosx).

`= -(lncosx)^2/2 + C`

Therefore, `int tanxln(cosx)dx = -(lncosx)^2/2+C` .

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