Evaluate the definite integral.?

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1 Answer
Nov 16, 2017

#int_0^1 6cos((pit)/2) dt = 12/pi#

Explanation:

#int_0^1 6cos((pit)/2) dt = 12/pi int_0^1 cos((pit)/2) d((pit)/2)#

#int_0^1 6cos((pit)/2) dt = 12/pi [sin((pit)/2)]_0^1#

#int_0^1 6cos((pit)/2) dt = 12/pi (sin(pi/2)-sin(0)) =12/pi#