integrate cos^2[x]dx from x=0 to pi/2?

Comments

• germano

  June 2021

  Hello,

  π/2

  ∫ cos²x dx =

  0

  let's apply the power-reduction formula cos²x = [1 + cos(2x)] /2:

  ∫ cos²x dx = ∫ {[1 + cos(2x)] /2} dx =

  ∫ [(1/2) + (1/2)cos(2x)] dx =

  (splitting into two integrals pulling constants out)

  (1/2) ∫ dx + (1/2) ∫ cos(2x) dx =

  (1/2)x + (1/2) [(1/2)sin(2x)] + C =

  (1/2)x + (1/4)sin(2x) + C (this is the antiderivative)

  once we have the antiderivative, let's plug in the bounds:

  π/2

  ∫ cos²x dx = {(1/2)(π/2) + (1/4)sin[2(π/2)]} - {(1/2)(0) + (1/4)sin[2(0)]} =

  0

  (π/4) + (1/4)sin(π) - 0 - (1/4)sin(0) =

  (π/4) + (1/4)(0) - (1/4)(0) =

  (π/4) + 0 = π/4

  the right answer is: π/4

  I hope it helps