# What's the integral of #int (tanx)ln(cosx)dx #?

Use substitution with

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To integrate ( \int \tan(x) \ln(\cos(x)) , dx ), you can use integration by parts. Let ( u = \ln(\cos(x)) ) and ( dv = \tan(x) , dx ).

Then, ( du = -\frac{\sin(x)}{\cos(x)} , dx ) and ( v = -\ln|\cos(x)| ).

Applying the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

yields:

[ \int \tan(x) \ln(\cos(x)) , dx = -\ln(\cos(x))\tan(x) - \int (-\ln|\cos(x)|) \frac{\sin(x)}{\cos(x)} , dx ]

[ = -\ln(\cos(x))\tan(x) + \int \ln(\cos(x)) \frac{\sin(x)}{\cos(x)} , dx ]

Now, integrate ( \int \ln(\cos(x)) \frac{\sin(x)}{\cos(x)} , dx ) using substitution.

Let ( t = \cos(x) ), then ( dt = -\sin(x) , dx ).

Substituting, the integral becomes:

[ -\int \ln(t) , dt ]

This is a standard integral, which is:

[ = -t\ln(t) + t + C ]

Substitute back ( t = \cos(x) ):

[ = -\cos(x)\ln(\cos(x)) + \cos(x) + C ]

So, the final integral is:

[ \int \tan(x) \ln(\cos(x)) , dx = -\ln(\cos(x))\tan(x) - \cos(x)\ln(\cos(x)) + \cos(x) + C ]

where ( C ) is the constant of integration.

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