## Appendix A  
The coefficient \(\mathrm{a}_{0}\) for the dc term in Equations (12) and (13) will be investigated. Applying Equation (11) to Equation (13), then  
\[a_{0}= \tag{1}\]  
The constant \(\gamma_{2}\) in Equation (11) is located within the integral region [\(-\)T/2, T/2], and therefore, the terms cancel each other out. By accounting for the configuration of the sine function in the first and the third term on the right hand side, the following result can be obtained:  
\[\therefore a_{0}=\frac{2}{T}\left[\int_{-T/4}^{T/4}\{\gamma_{1}(1+sin\,\omega t^{ \prime})^{x+2}-\gamma_{1}(1+sin\,\omega t^{\prime})^{x+2}\}dt\right]=0 \tag{2}\]  
From Equation (2), it can be confirmed that there is no dc term in the variation force that occurs during harmonic behavior. Next, the coefficient \(\mathrm{a}_{1}\) for the first harmonic component will be investigated. Similar to the case of \(\mathrm{a}_{0}\), applying Equation (11) to Equation (13) produces the following:  
\[a_{1}=\frac{2}{T}\left[ \begin{array}{c}\int_{-T/4}^{T/4}\{\gamma_{1}(1-sin \,\omega t)^{x+2}+\gamma_{2}\}\cdot cos\,\omega tdt+\\ \int_{-T/4}^{T/4}\{\gamma_{1}(1+sin\,\omega t)^{x+2}-\gamma_{2}\}\cdot cos\, \omega tdt+\\ \int_{T/4}^{T/2}\{-\gamma_{1}(1-sin\,\omega t)^{x+2}+\gamma_{2}\}\cdot cos\, \omega tdt\end{array} \right] \tag{3}\]  
Similar to the case of Equation (3), if taking the sine and cosine function into account, the terms in the first and the third from the top of the right hand side of Equation (3) can be transformed:  
\[a_{1}=\frac{2}{T}\left[ \begin{array}{c}\int_{0}^{T/4}\{\gamma_{1}(1+sin\, \omega t)^{x+2}-\gamma_{2}\}\cdot cos\,\omega tdt+\\ \int_{-T/4}^{T/4}\{\gamma_{1}(1+sin\,\omega t)^{x+2}-\gamma_{2}\}\cdot cos\, \omega tdt+\\ \int_{-T/4}^{T/4}\{\gamma_{1}(1+sin\,\omega t)^{x+2}-\gamma_{2}\}\cdot cos\, \omega tdt\end{array} \right] \tag{4}\]  
Re-arranging Equation (4) within the integral region leads to the following:  
\[a_{1}=\frac{4}{T}\left[\int_{-T/4}^{T/4}\gamma_{1}(1+sin\,\omega t)^{x+2}\, cos\,\omega tdt-\int_{-T/4}^{T/4}\gamma_{2}\,cos\,\omega tdt\right] \tag{5}\]  
The first term on the right hand side of Equation (5) can be obtained by a change of variable. If transforming \(sin\,\omega t\) to the variable \(z\), then the integral region will be transformed to [\(-\)1, 1]. Eventually, the result of the first term will be as follows:  
\[\int_{-1}^{1}\gamma_{1}(1+z)^{x+2}dz=\frac{1}{\omega}\frac{\gamma_{1}}{\chi+ 3}2^{x+3} \tag{6}\]  
The second term in Equation (5) can be obtained without difficulty. Therefore, the coefficient \(\mathrm{a}_{1}\) is  
\[\therefore a_{1}=\frac{4}{\pi}\left(\frac{\gamma_{1}}{\chi+3}2^{x+2}-\gamma_{2}\right) \tag{7}\]  
The coefficient \(\mathrm{b}_{1}\) of the second harmonic component of the base component will be investigated. Similar to the previous case of applying Equation (11) to Equation (13), by taking the configurations ofthe sine and cosine functions into account and transforming the integral region, the following result can be obtained:  
\[b_{1}=\frac{4}{T}\Biggl{[}\int_{-T/4}^{T/4}\left\{\gamma_{1}(1+sin\,\omega t)^{ \chi+2}-\gamma_{2}\right]sin\,\omega tdt\right] \tag{10}\]  
Because the sine function is an odd function, Equation (10) can be transformed as follows:  
\[b_{1}=\frac{4}{T}\int_{-T/4}^{T/4}f_{b}(t)dt\text{ where }f_{b}(t)=\gamma_{1}(1+ sin\,\omega t)^{\chi+2}\,sin\,\omega t \tag{11}\]  
Equation (11) cannot be applied to variable change method, and thus, [PERSON]'s series is used to obtain \(f_{b}\)(t) in Equation (11):  
\[p_{b}(t)=f_{b}(c)+f_{b}^{\prime}(c)(t-c)+\frac{f_{b}^{\prime\prime}(c)}{2!}(t- c)^{2}+\frac{f_{b}^{\prime\prime\prime}(c)}{3!}(t-c)^{3}+\cdots \tag{12}\]  
In Equation (12), \(p_{b}\) and \(c\) represent the Taylor polynomial and the center point of the Taylor series, respectively. The prime indicates differentiation with respect to time. In this study, up to the third terms in Equation (12) were accounted.