Appendix C

Integral Expressions for the Triangle Wave Example

This appendix solves two integral expressions for the triangle wave example:

1. \(\displaystyle\frac{4A}{T^2} \left( \color{red}{\int\limits_0^{T/2} t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {\ -\ } \color{green}{\int\limits_{-T/2}^0 t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {}\right)\)
2. \(\displaystyle\frac{4A}{T^2} \left( \color{red}{\int\limits_0^{T/2} t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {\ -\ } \color{green}{\int\limits_{-T/2}^0 t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {}\right)\)

The integral of type \(\displaystyle\int t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt\) or \(\displaystyle\int t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt\) can be solved using the integration by parts method.

Integration by parts reminder

The product rule (or Leibniz rule) is a formula used to find the derivatives of products of two or more functions:

\[(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\]

Integrate both parts:

\[\int \left((f(x)g(x))'\right)dx = \int \left(f'(x)g(x) + f(x)g'(x)\right)dx\]

\[f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx\]

\[\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx\]

Let’s make a couple of substitutions:

• \(u = f(x)\), \(du = f'(x)dx\)
• \(v = g(x)\), \(dv = g'(x)dx\)

This results in the integration by parts formula:

\[\int u\,dv = uv - \int v\,du\]

For a definite integral:

\[\int_a^b u\,dv = \left[ uv \right]_a^b - \int_a^b v\,du\]

Solving \(\displaystyle\int t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt\)

Let:

• \(u = t\), \(dv = \cos\left(\dfrac{2\pi m}{T} t\right)dt\)
• \(du = dt\), \(v = \dfrac{T}{2\pi m} \sin\left(\dfrac{2\pi m}{T} t\right)\)

\[\displaystyle\int\limits_a^b t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt = \int\limits_a^b u\,dv\]

Using \(\displaystyle\int\limits_a^b u\,dv = \left[uv\right]_a^b - \int\limits_a^b v\,du\), we find:

\[\displaystyle\int\limits_a^b t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt = \left[ t \cdot \frac{T}{2\pi m} \sin\left(\frac{2\pi m}{T} t\right) \right]_a^b - \frac{T}{2\pi m} \int\limits_a^b \sin\left(\frac{2\pi m}{T} t\right)dt\]

\[= \left[ t \cdot \frac{T}{2\pi m} \sin\left(\frac{2\pi m}{T} t\right) \right]_a^b + \frac{T}{2\pi m} \left[ \frac{T}{2\pi m} \cos\left(\frac{2\pi m}{T} t\right) \right]_a^b\]

\[= \frac{T}{2\pi m} \left[ t \cdot \sin\left(\frac{2\pi m}{T} t\right) \right]_a^b + \left(\frac{T}{2\pi m}\right)^2 \left[ \cos\left(\frac{2\pi m}{T} t\right) \right]_a^b\]

First Integral Term:

\[\color{red}{\displaystyle\int\limits_0^{T/2} t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt = \frac{T}{2\pi m} \left[ t \cdot \sin\left(\frac{2\pi m}{T} t\right) \right]_0^{T/2} + \left(\frac{T}{2\pi m}\right)^2 \left[ \cos\left(\frac{2\pi m}{T} t\right) \right]_0^{T/2}}\]

\[\color{red}{= \frac{T}{2\pi m} \left(\frac{T}{2} \sin(\pi m) - 0\right) + \left(\frac{T}{2\pi m}\right)^2 \left(\cos(\pi m) - \cos(0)\right)}\]

Since \(\sin(\pi m) = 0\):

\[\color{red}{= \left(\frac{T}{2\pi m}\right)^2 \left(\cos(\pi m) - 1\right)}\]

Second Integral Term:

\[\color{green}{\displaystyle\int\limits_{-T/2}^0 t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt = \frac{T}{2\pi m} \left[ t \cdot \sin\left(\frac{2\pi m}{T} t\right) \right]_{-T/2}^0 + \left(\frac{T}{2\pi m}\right)^2 \left[ \cos\left(\frac{2\pi m}{T} t\right) \right]_{-T/2}^0}\]

\[\color{green}{= \frac{T}{2\pi m} \left(0 + \frac{T}{2} \sin(-\pi m)\right) + \left(\frac{T}{2\pi m}\right)^2 \left(\cos(0) - \cos(-\pi m)\right)}\]

Since \(\sin(-\pi m) = 0\) and \(\cos(-\pi m) = \cos(\pi m)\):

\[\color{green}{= \left(\frac{T}{2\pi m}\right)^2 \left(1 - \cos(\pi m)\right)}\]

Combined Expression:

\[\frac{4A}{T^2} \left( \color{red}{\int\limits_0^{T/2} t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {\ +\ } \color{green}{\int\limits_{-T/2}^0 t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {}\right)\]

\[= \frac{4A}{T^2} \left( \color{red}{\left(\frac{T}{2\pi m}\right)^2 (\cos(\pi m) - 1)} \color{#222832} {\ -\ } \color{green}{\left(\frac{T}{2\pi m}\right)^2 (1 - \cos(\pi m))}\color{#222832} {} \right)\]

\[= \frac{2A}{(\pi m)^2} (\cos(\pi m) - 1)\]

Solving \(\displaystyle\int t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt\)

Let:

• \(u = t\), \(dv = \sin\left(\dfrac{2\pi m}{T} t\right)dt\)
• \(du = dt\), \(v = -\dfrac{T}{2\pi m} \cos\left(\dfrac{2\pi m}{T} t\right)\)

\[\displaystyle\int\limits_a^b t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt = \int\limits_a^b u\,dv\]

Using \(\displaystyle\int\limits_a^b u\,dv = \left[uv\right]_a^b - \int\limits_a^b v\,du\), we find:

\[\displaystyle\int\limits_a^b t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt = \left[ -t \cdot \frac{T}{2\pi m} \cos\left(\frac{2\pi m}{T} t\right) \right]_a^b + \frac{T}{2\pi m} \int\limits_a^b \cos\left(\frac{2\pi m}{T} t\right)dt\]

\[= \left[ -t \cdot \frac{T}{2\pi m} \cos\left(\frac{2\pi m}{T} t\right) \right]_a^b + \frac{T}{2\pi m} \left[ \frac{T}{2\pi m} \sin\left(\frac{2\pi m}{T} t\right) \right]_a^b\]

\[= -\frac{T}{2\pi m} \left[ t \cdot \cos\left(\frac{2\pi m}{T} t\right) \right]_a^b + \left(\frac{T}{2\pi m}\right)^2 \left[ \sin\left(\frac{2\pi m}{T} t\right) \right]_a^b\]

First Integral Term:

\[\color{red}{\displaystyle\int\limits_0^{T/2} t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt = -\frac{T}{2\pi m} \left[ t \cdot \cos\left(\frac{2\pi m}{T} t\right) \right]_0^{T/2} + \left(\frac{T}{2\pi m}\right)^2 \left[ \sin\left(\frac{2\pi m}{T} t\right) \right]_0^{T/2}}\]

\[\color{red}{= -\frac{T}{2\pi m} \left(\frac{T}{2} \cos(\pi m) - 0\right) + \left(\frac{T}{2\pi m}\right)^2 \left(\sin(\pi m) - \sin(0)\right)}\]

\[\color{red}{= -\frac{T^2}{4\pi m} \cos(\pi m)}\]

Second Integral Term:

\[\color{green}{\displaystyle\int\limits_{-T/2}^0 t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt = -\frac{T}{2\pi m} \left[ t \cdot \cos\left(\frac{2\pi m}{T} t\right) \right]_{-T/2}^0 + \left(\frac{T}{2\pi m}\right)^2 \left[ \sin\left(\frac{2\pi m}{T} t\right) \right]_{-T/2}^0}\]

\[= \color{green}{\frac{T}{2\pi m} \left(0 - \frac{T}{2} \cos(-\pi m)\right) + \left(\frac{T}{2\pi m}\right)^2 \left(\sin(0) - \sin(-\pi m)\right)}\]

\[= \color{green}{-\frac{T^2}{4\pi m} \cos(\pi m)}\]

Combined Expression:

\[\frac{4A}{T^2} \left( \color{red}{\int\limits_0^{T/2} t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {\ -\ } \color{green}{\int\limits_{-T/2}^0 t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {} \right)\]

\[= \frac{4A}{T^2} \left( \color{red}{ -\left(\frac{T}{2\pi m}\right)^2 \cos(\pi m)} \color{#222832} {\ +\ } \color{green}{\left(\frac{T}{2\pi m}\right)^2 \cos(\pi m)} \color{#222832} {}\right) = 0\]

Final Results:

1. \(\displaystyle \frac{4A}{T^2} \left( \color{red}{\int\limits_0^{T/2} t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {\ -\ } \color{green}{\int\limits_{-T/2}^0 t \cdot \cos\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {} \right) = \frac{2A}{(\pi m)^2} \left(\cos(\pi m) - 1 \right) \)
2. \(\displaystyle \frac{4A}{T^2} \left( \color{red}{\int\limits_0^{T/2} t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {\ -\ } \color{green}{\int\limits_{-T/2}^0 t \cdot \sin\left(\frac{2\pi m}{T} t\right)dt} \color{#222832} {} \right) = 0\)