Why integrating sin^2(wt) gives the half-period?

[blanxink]

In an exercise with included solution I can't understand how integrating sin^2(ωt) gives T(period)/2

$\int$sin^2(ωt)dt = Period/2

I posted the whole problem below, because I had more doubts, but understood them typing up the problem.

I appreciate any help.

YOU CAN SKIP THE PROBLEM

The probelm

A circular coil, r=10 and Ω=1.5Ω, rotates around its diameter with a constant ω0 in a uniform and constant magnetic field B that forms an angle of θ=∏/3 with the axis of rotation of the coil.
Knowing that the maximum current Imax=0.15A, and that the component of B parallel to the axis of rotation is Bparall=1.0T, find
1) intensity of B
2) angular velocity ω0 of the coil
3) energy needed per rotation to keep the angular velocity ω0

The solution included on the book
my problem in in red

1) B=Bparallel/cosθ=2Bparallel=2.0T
I did the same, so no problem here.

Bperp(responsible for the current in the coil)=Bparallel*tanθ=Bparallel√3

2) fem=-dphi/dt=∏*r^2*ω*Bperp*sin(ωt), max fem when sin(ωt)=1,
ω0=R*I/(∏*r^2*Bperp)

3)To get the work, i integrate over one turn, so over the period T=2∏/ω
W= $\int$R*I^2*dt = R*Imax^2$\int$sin^2(ωt)dt = R*I^2*Period/2
I don't get how do you integrate sin^2(wt) and get T/2?

 

[gneill]

In an exercise with included solution I can't understand how integrating sin^2(ωt) gives T(period)/2

$\int$sin^2(ωt)dt = Period/2

I posted the whole problem below, because I had more doubts, but understood them typing up the problem.

I appreciate any help.

YOU CAN SKIP THE PROBLEM

The probelm

A circular coil, r=10 and Ω=1.5Ω, rotates around its diameter with a constant ω0 in a uniform and constant magnetic field B that forms an angle of θ=∏/3 with the axis of rotation of the coil.
Knowing that the maximum current Imax=0.15A, and that the component of B parallel to the axis of rotation is Bparall=1.0T, find
1) intensity of B
2) angular velocity ω0 of the coil
3) energy needed per rotation to keep the angular velocity ω0

The solution included on the book
my problem in in red

1) B=Bparallel/cosθ=2Bparallel=2.0T
I did the same, so no problem here.

Bperp(responsible for the current in the coil)=Bparallel*tanθ=Bparallel√3

2) fem=-dphi/dt=∏*r^2*ω*Bperp*sin(ωt), max fem when sin(ωt)=1,
ω0=R*I/(∏*r^2*Bperp)

3)To get the work, i integrate over one turn, so over the period T=2∏/ω
W= $\int$R*I^2*dt = R*Imax^2$\int$sin^2(ωt)dt = R*I^2*Period/2
I don't get how do you integrate sin^2(wt) and get T/2?

How is ω related to T?

 

[Simon Bridge]

How is ω related to T?

$\omega = \frac{2\pi}{T}$ ... like you said above.
Your limits of integration are 0-2π?


(note: π is the lower case of ∏.)

 

[D H]

I don't get how do you integrate sin^2(wt) and get T/2?

Do you know what the integral of $\sin^2x$ is? If you don't, use the cosine double angle formula $\cos 2x = \cos^2x - \sin^2x$ and the Pythagorean identity $\sin^2x + \cos^2x = 1$ to express $\sin^2x$ in terms of $\cos 2x$. From there it should be easy sailing.