Answer :

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


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