∫√(v^2 sin^2(2π/T t)) dt
Looks to me like just
∫|v sin(2π/T t)| dt
The absolute value throws in a wrinkle, but the integral is simple enough.
indefinite integral : (sqrt(V^2sin^2(2pi/T*t))) dt
2 answers
with this problem, you can see there are two different function : v and sin
so what you need to do is set u=v^2 => du = 2vdv and dv=sin^2... => v=
then apply the familiar formula : uv-∫vdu but i think you will have to do twice time for integral.
so what you need to do is set u=v^2 => du = 2vdv and dv=sin^2... => v=
then apply the familiar formula : uv-∫vdu but i think you will have to do twice time for integral.