- #1

Const@ntine

- 285

- 18

_{0}

^{T}sin

^{2}(ωt) dt

The whole thing: 1/T∫

_{0}

^{T}sin

^{2}(ωt) dt = 1/T(t/2 + sin2ωt/2ω)|

^{T}

_{0}= 1/2

I didn't remember how to integrate that, so I went back to check my notes, and look at it at Wolfram or some other sites. But the problem is, I get different results. Same for the other sites I checked.

Wolfram: http://www.wolframalpha.com/input/?i=integrate+sin^2(ω*t)

Cymath (it's a simpler form, but the theory is the same): https://www.cymath.com/answer?q=int(sin(x)^2,x)

So I went to try my hand at it:

∫

_{0}

^{T}sin

^{2}(ωt) dt

u = ωt <=> du = ωdt

sin

^{2}(x) = 1/2 - cos(2x)/2

1/ω∫

_{0}

^{T}1/2 - cos2u/2 du = 1/ω [ ∫

_{0}

^{T}1/2 du - ∫

_{0}

^{T}cos2u/2 du]

We break that into two integrals:

1/ω∫

_{0}

^{T}1/2 du = 1/ω (u/2)|

^{T}

_{0}= 1/ω (ωt/2)|

^{T}

_{0}= T/2

1/ω∫

_{0}

^{T}cos2u/2 du

k = 2u <=> dk = 2du

1/ω∫

_{0}

^{T}cosk/4 dk = 1/4ω∫

_{0}

^{T}cosk dk = 1/4ω(sink)|

^{T}

_{0}= 1/4ω(sin2u)|

^{T}

_{0}= 1/4ω(sin2ωt)|

^{T}

_{0}= sin(2ωT)/4ω

But, ω = 2π/T so sin(2ωT)/4ω = 0

In the end, we have: 1/T*(T/2 - 0) = 1/2

The end result is the same, but I wonder which version is correct:

Book: 1/T∫

_{0}

^{T}sin

^{2}(ωt) dt = 1/T(t/2 + sin2ωt/2ω)|

^{T}

_{0}

Other: 1/T∫

_{0}

^{T}sin

^{2}(ωt) dt = 1/T(t/2 - sin2ωt/4ω)|

^{T}

_{0}

Any help is appreciated!

PS: I know that I also have to change the T & 0 when changing the integrating factor (t, then u, then k), but it's been a few years since I've done that, and have forgotten how exactly, so I justkept them the same and reverted the u & k back to their original t-related forms when the time came to find the end result.