Question #8c45d

1 Answer
Leland Adriano Alejandro
Nov 30, 2016

#int_0^1 2 cos ((pit)/2) dt=4/pi#

Explanation:

Provide the preparation #pi/2# inside the integral and #2/pi# outside the integral.

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

#int_0^1 2 cos ((pit)/2) dt=(4/pi)[sin ((pit)/2)]_0^1#

#int_0^1 2 cos ((pit)/2) dt=4/pi[sin ((pi/2)(1))-sin ((pi/2)(0))]#

#int_0^1 2 cos ((pit)/2) dt=4/pi#

God bless....I hope the explanation is useful.

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