Question #7c8b2

1 Answer
Samuel A.
Feb 7, 2018

#pi/2#

Explanation:

#I = int_0 ^(pi/2) sin^(0)(x)dx#
You could subst #sin^0 (x)# with 1, but you must take care when #x=0#. But, in integration we just need to have a well defined value near to the bounds of integration.
#I = lim_(a->0^+) int_a^(pi/2)1dx=lim_(a->0^+) (pi/2-a)=pi/2#

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