# Evaluate the indefinite integral. integrate of cot(x)ln(sin(x))dx

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You should evaluate the integral using the following substitution such that:

`sin x = t => cos x dx = dt`

Changing the variable yields:

`int ln(sin x) cos x dx = int ln t dt`

You should use integration by parts formula such that:

`int udv = uv - int vdu`

Reasoning by analogy yields:

`u = ln t => du = (dt)/t`

`dv = dt => v =` `t`

`int ln t dt = t*ln t - int t*(1/t)dt`

`int ln t dt = t*ln t - int dt => int ln t dt = t*ln t - t + c`

Factoring out t yields:

`int ln t dt = t*(ln t - 1) + c`

Substituting back `sin x` for t yields:

`int ln(sin x) cos x dx = sin x*(ln(sin x) - 1) + c`

Hence, evaluating the given indefinite integral , using two methods of integration, yields `int ln(sin x) cos x dx = sin x*(ln(sin x) - 1) + c.`

Let;

`t = lnsinx`

Then;

`(dt)/dx = 1/sinx*cosx = cotx`

`dt = cotxdx`

`int cotxlnsinxdx`

`= int lnsinxcotxdx`

`= int tdt`

`= t^2/2+C` where C is a constant

`= 1/2(lnsinx)^2+C`

`= lnsinx+C`

`int cotxlnsinxdx = lnsinx+C`