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`

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now