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

well the first pattern to note is this

And so

And thus

Or you can try find a sub

This one works well:

...because the integral becomes

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To integrate ( (\tan{x})\ln{(\cos{x})} ) with respect to ( x ), 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, the integral becomes:

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

This second integral can be solved using substitution method. Let ( t = \cos{x} ), then ( dt = -\sin{x} , dx ). Substituting into the integral, we get:

[ \int \ln{(\cos{x})} \cdot \frac{\sin{x}}{\cos{x}} , dx = -\int \ln{t} , dt = -t\ln{t} + t + C ]

Substitute ( t = \cos{x} ) back in, we get:

[ -\cos{x}\ln{|\cos{x}|} + \cos{x} + C ]

So, the final answer is:

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

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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