How do you integrate #int cos((2x)/3)# from [0, pi/2]?

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Answer 1 · 2017-02-13T22:27:00

#int_0^(pi/2) cos((2x)/3)dx = 3/2 int_0^(pi/2) cos((2x)/3)d((2x)/3) = 3/2 [sin((2x)/3)]_0^(pi/2) = 3/2( sin (pi/3) -sin 0) = (3sqrt(3))/4#