$$$i n t \sin^{2}{\left(\frac{\pi x}{l} \right)}$$$ 對 $$$x$$$ 的積分

求$$$\int i n t \sin^{2}{\left(\frac{\pi x}{l} \right)}\, dx$$$。

套用降冪公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，令 $$$\alpha=\frac{\pi x}{l}$$$:

$${\color{red}{\int{i n t \sin^{2}{\left(\frac{\pi x}{l} \right)} d x}}} = {\color{red}{\int{\frac{i n t \left(1 - \cos{\left(\frac{2 \pi x}{l} \right)}\right)}{2} d x}}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = i n t \left(1 - \cos{\left(\frac{2 \pi x}{l} \right)}\right)$$$：

$${\color{red}{\int{\frac{i n t \left(1 - \cos{\left(\frac{2 \pi x}{l} \right)}\right)}{2} d x}}} = {\color{red}{\left(\frac{\int{i n t \left(1 - \cos{\left(\frac{2 \pi x}{l} \right)}\right) d x}}{2}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{i n t \left(1 - \cos{\left(\frac{2 \pi x}{l} \right)}\right) d x}}}}{2} = \frac{{\color{red}{\int{\left(- i n t \cos{\left(\frac{2 \pi x}{l} \right)} + i n t\right)d x}}}}{2}$$

逐項積分：

$$\frac{{\color{red}{\int{\left(- i n t \cos{\left(\frac{2 \pi x}{l} \right)} + i n t\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{i n t d x} - \int{i n t \cos{\left(\frac{2 \pi x}{l} \right)} d x}\right)}}}{2}$$

配合 $$$c=i n t$$$，應用常數法則 $$$\int c\, dx = c x$$$：

$$- \frac{\int{i n t \cos{\left(\frac{2 \pi x}{l} \right)} d x}}{2} + \frac{{\color{red}{\int{i n t d x}}}}{2} = - \frac{\int{i n t \cos{\left(\frac{2 \pi x}{l} \right)} d x}}{2} + \frac{{\color{red}{i n t x}}}{2}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=i n t$$$ 與 $$$f{\left(x \right)} = \cos{\left(\frac{2 \pi x}{l} \right)}$$$：

$$\frac{i n t x}{2} - \frac{{\color{red}{\int{i n t \cos{\left(\frac{2 \pi x}{l} \right)} d x}}}}{2} = \frac{i n t x}{2} - \frac{{\color{red}{i n t \int{\cos{\left(\frac{2 \pi x}{l} \right)} d x}}}}{2}$$

令 $$$u=\frac{2 \pi x}{l}$$$。

則 $$$du=\left(\frac{2 \pi x}{l}\right)^{\prime }dx = \frac{2 \pi}{l} dx$$$ (步驟見»)，並可得 $$$dx = \frac{l du}{2 \pi}$$$。

該積分變為

$$\frac{i n t x}{2} - \frac{i n t {\color{red}{\int{\cos{\left(\frac{2 \pi x}{l} \right)} d x}}}}{2} = \frac{i n t x}{2} - \frac{i n t {\color{red}{\int{\frac{l \cos{\left(u \right)}}{2 \pi} d u}}}}{2}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{l}{2 \pi}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

$$\frac{i n t x}{2} - \frac{i n t {\color{red}{\int{\frac{l \cos{\left(u \right)}}{2 \pi} d u}}}}{2} = \frac{i n t x}{2} - \frac{i n t {\color{red}{\left(\frac{l \int{\cos{\left(u \right)} d u}}{2 \pi}\right)}}}{2}$$

餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：

$$- \frac{i l n t {\color{red}{\int{\cos{\left(u \right)} d u}}}}{4 \pi} + \frac{i n t x}{2} = - \frac{i l n t {\color{red}{\sin{\left(u \right)}}}}{4 \pi} + \frac{i n t x}{2}$$

回顧一下 $$$u=\frac{2 \pi x}{l}$$$：

$$- \frac{i l n t \sin{\left({\color{red}{u}} \right)}}{4 \pi} + \frac{i n t x}{2} = - \frac{i l n t \sin{\left({\color{red}{\left(\frac{2 \pi x}{l}\right)}} \right)}}{4 \pi} + \frac{i n t x}{2}$$

因此，

$$\int{i n t \sin^{2}{\left(\frac{\pi x}{l} \right)} d x} = - \frac{i l n t \sin{\left(\frac{2 \pi x}{l} \right)}}{4 \pi} + \frac{i n t x}{2}$$

化簡：

$$\int{i n t \sin^{2}{\left(\frac{\pi x}{l} \right)} d x} = \frac{i n t \left(- l \sin{\left(\frac{2 \pi x}{l} \right)} + 2 \pi x\right)}{4 \pi}$$

加上積分常數：

$$\int{i n t \sin^{2}{\left(\frac{\pi x}{l} \right)} d x} = \frac{i n t \left(- l \sin{\left(\frac{2 \pi x}{l} \right)} + 2 \pi x\right)}{4 \pi}+C$$

$$$\int i n t \sin^{2}{\left(\frac{\pi x}{l} \right)}\, dx = \frac{i n t \left(- l \sin{\left(\frac{2 \pi x}{l} \right)} + 2 \pi x\right)}{4 \pi} + C$$$A