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Given, ∫cos^-1(sin x ) dx

Let us consider, ∫cos^-1dx

We know that, f(x).g(x) dx=g(x) dx-[f^I(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