integral square root of sin to the power of negative 1 end exponent open parentheses cos x close parentheses end root d x space equals space ?
e x p l a i n
give answer

Asked by modi72879 | 11th Nov, 2017, 06:36: PM

Expert Answer:

begin mathsize 16px style sin to the power of negative 1 end exponent left parenthesis cosx right parenthesis equals straight pi over 2 minus cos to the power of negative 1 end exponent left parenthesis cosx right parenthesis
equals straight pi over 2 minus straight x
integral space becomes
integral square root of straight pi over 2 minus straight x end root space dx
put space straight pi over 2 minus straight x equals straight t
and space solve end style

Answered by Arun | 7th Dec, 2017, 12:32: PM