9.8 Show that lebi to+Tendo to aburtlar V2T 2 V cos² (wt + o) dt / onsupor -

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Circuits 1 HW 6 Q3
**Exercise 9.8**

Show that:

\[
\int_{t_0}^{t_0 + T} V_m^2 \cos^2(\omega t + \phi) \, dt = \frac{V_m^2 T}{2}
\]

This equation involves evaluating the integral of the square of a cosine function over one period \( T \). The expression \( V_m^2 \cos^2(\omega t + \phi) \) represents a squared sinusoidal function, where:
- \( V_m \) is the maximum amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase angle,
- \( t \) is time,
- \( T \) is the period of the cosine function.

The outcome of the integral is shown to be one-half of the product of the square of the amplitude and the period.
Transcribed Image Text:**Exercise 9.8** Show that: \[ \int_{t_0}^{t_0 + T} V_m^2 \cos^2(\omega t + \phi) \, dt = \frac{V_m^2 T}{2} \] This equation involves evaluating the integral of the square of a cosine function over one period \( T \). The expression \( V_m^2 \cos^2(\omega t + \phi) \) represents a squared sinusoidal function, where: - \( V_m \) is the maximum amplitude, - \( \omega \) is the angular frequency, - \( \phi \) is the phase angle, - \( t \) is time, - \( T \) is the period of the cosine function. The outcome of the integral is shown to be one-half of the product of the square of the amplitude and the period.
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