More evaluate the integrals help?

Question

∫cosx√(sinx-1)dx

 

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Answer 1

Given integral ∫cosx√(sinx-1)dx

= ∫cosx√(sinx-1)dx

Let  sin x  - 1= t

cos x dx = dt , then

= ∫√t dt

= (t ^ 1/2+1)/ (1/2+1) + c

= (t^ 3/2)/3/2 + c

= (2/3)t^ 3/2 + c

But t = sin x - 1

= (2/3)(sin x - 1)^ 3/2 + c