Calculate the integral: integral from 0 to (T/2) sin ((2

Calculate the integral: $\int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}-\varphi_0\right) dt$.

We will use the Newton-Leibniz formula after finding the antiderivative.

Step 1: Find the antiderivative

We need to find the antiderivative of $\sin\left(\frac{2\pi t}{T}-\varphi_0\right)$ with respect to $t$.

Let $u = \frac{2\pi t}{T}-\varphi_0$, so $\frac{du}{dt} = \frac{2\pi}{T}$, which means $dt = \frac{T}{2\pi} \, du$.

We know that $\int \sin(u) \, du = -\cos(u) + C$, therefore:

\[\int \sin\left(\frac{2\pi t}{T}-\varphi_0\right) dt = -\frac{T}{2\pi} \cos\left(\frac{2\pi t}{T}-\varphi_0\right) + C\]

Step 2: Apply the Newton-Leibniz formula

\[\int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}-\varphi_0\right) dt = \left[-\frac{T}{2\pi} \cos\left(\frac{2\pi t}{T}-\varphi_0\right)\right]_0^{\frac{T}{2}}\]

Step 3: Evaluate at the upper limit $t = \frac{T}{2}$

\[F\left(\frac{T}{2}\right) = -\frac{T}{2\pi} \cos\left(\pi - \varphi_0\right)\]

Using the identity $\cos(\pi - \varphi_0) = -\cos(\varphi_0)$:

\[= -\frac{T}{2\pi} \cdot (-\cos(\varphi_0)) = \frac{T}{2\pi} \cos(\varphi_0)\]

Step 4: Evaluate at the lower limit $t = 0$

\[F(0) = -\frac{T}{2\pi} \cos(-\varphi_0)\]

Using the identity $\cos(-\varphi_0) = \cos(\varphi_0)$ (cosine is even):

\[= -\frac{T}{2\pi} \cos(\varphi_0)\]

Step 5: Calculate the definite integral

\[\int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}-\varphi_0\right) dt = F\left(\frac{T}{2}\right) - F(0) = \frac{T}{2\pi} \cos(\varphi_0) - \left(-\frac{T}{2\pi} \cos(\varphi_0)\right) = \frac{2T}{2\pi} \cos(\varphi_0) = \frac{T}{\pi} \cos(\varphi_0)\]

Physical interpretation:

This integral appears frequently in the analysis of periodic signals (such as in electrical engineering or physics): $T$ represents the period of oscillation, $\frac{2\pi t}{T}$ is the angular frequency times time, $\varphi_0$ is the phase shift, and the integral represents the average value of the sine function over half a period (scaled by $T$).

Special cases:

• If $\varphi_0 = 0$: $\int_0^{T/2} \sin\left(\frac{2\pi t}{T}\right) dt = \frac{T}{\pi} \cos(0) = \frac{T}{\pi}$
• If $\varphi_0 = \frac{\pi}{2}$: $\int_0^{T/2} \sin\left(\frac{2\pi t}{T}-\frac{\pi}{2}\right) dt = \frac{T}{\pi} \cos\left(\frac{\pi}{2}\right) = 0$. This makes sense since $\sin\left(\frac{2\pi t}{T}-\frac{\pi}{2}\right) = -\cos\left(\frac{2\pi t}{T}\right)$, and integrating cosine over half a period gives zero due to symmetry.
• If $\varphi_0 = \pi$: $\int_0^{T/2} \sin\left(\frac{2\pi t}{T}-\pi\right) dt = \frac{T}{\pi} \cos(\pi) = -\frac{T}{\pi}$

Final answer:

\[\boxed{\int_0^{\frac{T}{2}} \sin\left(\frac{2\pi t}{T}-\varphi_0\right) dt = \frac{T}{\pi} \cos(\varphi_0)}\]