4 Removing obsolete sentence
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Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2}) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2}) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2})) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2})) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2}) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2}) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2})) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2})) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

3 Added extra brackets [] possiblty redundant in this case but adds clarity.
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Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align}$$$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

2 Clarification for the different cases
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Any mixed product between two different sines will have an integral that results to 0. You can useExpanding the following equality for that:sum will result in

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align}$$

Like so:If \$n_1 = n_2 = n\$, then we simply get

Assume$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, thenwe can calculate the integral using the following equality:

$$\int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ \begin{align} &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} 2I_n^2 \int_0^T \sin^2(n\omega t + \gamma_n)dt$$$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

Any mixed product between two different sines will have an integral that results to 0. You can use the following equality for that:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

Like so:

Assume \$n_1 \neq n_2\$, then

$$\int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ \begin{align} &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} 2I_n^2 \int_0^T \sin^2(n\omega t + \gamma_n)dt$$

The last integral should yield 1/2, leading you to the final result.

Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2}) \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_1)\sin(n_2\omega t + \gamma_2)dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_1 - \gamma_2) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_1 - \gamma_2)) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_1 + \gamma_2)) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

The last integral should yield 1/2, leading you to the final result.

1
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