Integration of sine square theta from 0 to 2pi
sin^2θ = (1-cos2θ)/2
so, you want
∫[0,2π] 1/2 (1-cos2θ) dθ = 1/2 (θ - 1/2 sin2θ) [0,2π] = π
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
To find the integral of sin^2(theta) from 0 to 2pi, we can use the trigonometric identity:
sin^2(theta) = (1/2)*(1 - cos(2theta))
Now we can rewrite the integral as:
∫[0 to 2pi] sin^2(theta) d(theta) = ∫[0 to 2pi] (1/2)*(1 - cos(2theta)) d(theta)
Let's break down the integral into two parts:
First, we integrate the term (1/2) with respect to theta:
∫[0 to 2pi] (1/2) d(theta) = (1/2) * ∫[0 to 2pi] d(theta) = (1/2) * [theta]_[0 to 2pi] = (1/2) * (2pi - 0) = pi
Second, we integrate the term (-1/2)cos(2theta) with respect to theta:
∫[0 to 2pi] (-1/2)cos(2theta) d(theta)
Let's substitute u = 2theta, which means du = 2 d(theta):
∫[0 to pi] (-1/2)cos(u) * (1/2) du = (-1/4) * ∫[0 to pi] cos(u) du
Now we can integrate cos(u) with respect to u:
(-1/4) * sin(u) |_[0 to pi]
Substituting back u = 2theta:
(-1/4) * sin(2theta) |_[0 to 2pi]
Now, we can plug in the upper and lower limits of the integral:
(-1/4) * [sin(4pi) - sin(0)]
Since sin(0) = 0 and sin(4pi) = 0, we have:
(-1/4) * (0 - 0) = 0
Putting everything together, the definite integral of sin^2(theta) from 0 to 2pi is equal to pi + 0 = pi.