Ηelp for definite integral
$\displaystyle V^2=\frac{1}{2\pi}\left \{\int_{0}^{\pi} \eta \mu^2 (vt)d(vt)\right \}^{\frac{1}{2}}$
Calculating the integral gives us:
$\displaystyle \int_{0}^{\pi} \eta \mu^2 (vt)d(vt)\right = \eta \mu^2 \int_{0}^{\pi} (vt)d(vt) = \eta \mu^2\left[\frac{(vt)^2}{2}\right]_{0}^{\pi} = \eta \mu^2\left(\frac{\pi^2}{2}\right)$
The answer should be clear now.
Siron,
I' m sure that what you wrote me does not hold. It is not allowed to seperate sin^2 from (ut), because sinf(x) is a function.
I appreciate your taking the time to write to me but it would be better not to help, if you do not know.
M.
$\displaystyle u = \omega t$
$\displaystyle \int_0^\pi \sin^2{u} \, du$
$\displaystyle \frac{1}{2} \int_0^\pi 1 - \cos(2u) \, du$
$\displaystyle \frac{1}{2} \left[u - \frac{\sin(2u)}{2} \right]_0^\pi$
$\displaystyle \frac{1}{2} \left[\left(\pi - 0 \right)-\left(0-0\right)\right] = \frac{\pi}{2}$