It is given that
∫ sin x log(cos x) dx
Take cos x = t
So we get
– sin x dx = dt
It can be written as
∫ sin x log(cos x) dx = – ∫log t dt = – ∫1. log t dt
Consider first function as log t and second function as 1
By integrating w.r.t t
= – log t. t + ∫1/t dt
Again by integrating the second term
= – t log t + t + c
Now replace t as cos x
= t(- log t + 1) + c
We get
= cos x(1 – log (cos x)) + c
