Evaluate:
Given, ∫cos-1(sin x ) dx
Let us consider, ∫cos-1dx
We know that, f(x).g(x) dx=g(x) dx-[fI(x)] dx
By comparison, f(x) =cos-1x ; g(x)=1
( since, )
=x cos-1x – (1-x2)1/2 +c
Therefore,
Replace ‘x’ with :-
ð
=sinx.cos-1x (sinx) –cosx+c