indefinite integral : (sqrt(V^2sin^2(2pi/T*t))) dt

∫√(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.

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.

To find the indefinite integral of the function √(V^2 sin^2(2π/T*t)) dt, we can follow these steps:

Step 1: Simplify the Integral
Since sin^2(2π/T*t) is non-negative, we can remove the square root by squaring both sides of the equation.

√(V^2 sin^2(2π/T*t)) dt = V * sin(2π/T*t) dt

Step 2: Apply the Power Rule for Integration
The integral of sin(2π/T*t) dt can be found using the power rule for integration. The power rule states that the integral of x^n dx is equal to (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.

∫ sin(2π/T*t) dt = -(T/(2π)) * cos(2π/T*t) + C

Step 3: Determine the Final Result
Now, we substitute back the original expression to obtain the final result:

∫ √(V^2 sin^2(2π/T*t)) dt = ∫ V * sin(2π/T*t) dt = -(V*T/(2π)) * cos(2π/T*t) + C

Therefore, the indefinite integral of √(V^2 sin^2(2π/T*t)) dt is equal to -(V*T/(2π)) * cos(2π/T*t) + C, where C is the constant of integration.