Hi! I came across a proof in my physics textbook (amperage=wattage/area), and it contained this integration: ∫0 T sin2(ωt) dt The whole thing: 1/T∫0 T sin2(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 sin2(ωt) dt u = ωt <=> du = ωdt sin2(x) = 1/2 - cos(2x)/2 1/ω∫0 T 1/2 - cos2u/2 du = 1/ω [ ∫0 T1/2 du - ∫0 T cos2u/2 du] We break that into two integrals: 1/ω∫0 T1/2 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 sin2(ωt) dt = 1/T(t/2 + sin2ωt/2ω)|T 0 Other: 1/T∫0 T sin2(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.