Question
Integration of sine square theta from 0 to 2pi
Answers
Steve
sin^2θ = (1-cos2θ)/2
so, you want
∫[0,2π] 1/2 (1-cos2θ) dθ = 1/2 (θ - 1/2 sin2θ) [0,2π] = π
so, you want
∫[0,2π] 1/2 (1-cos2θ) dθ = 1/2 (θ - 1/2 sin2θ) [0,2π] = π
bobpursley
INT sin^2 T dT over t=0 to 2PI
use the half angle formulas sin^2(T)= (1-cos(2T))/2
Int .5*(T-cos(2T)dt
= .5T - .25sin2T over limits
= .5(2PI)-.25sin(4PI)-.5*0+.25(sin(0)=PI
use the half angle formulas sin^2(T)= (1-cos(2T))/2
Int .5*(T-cos(2T)dt
= .5T - .25sin2T over limits
= .5(2PI)-.25sin(4PI)-.5*0+.25(sin(0)=PI