Question

Question #8c45d

Answer 1

`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.