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

nancy

Can someone please help me with this problem... my professor worked it today and got 0...but he said he didn't think it was right... I just want to double check? Thank you so much!

germano

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