$$$\sin^{2}{\left(\frac{\pi m x}{a} \right)}$$$$$$x$$$ 的積分

此計算器會求出 $$$\sin^{2}{\left(\frac{\pi m x}{a} \right)}$$$$$$x$$$ 的不定積分/原函數,並顯示步驟。

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您的輸入

$$$\int \sin^{2}{\left(\frac{\pi m x}{a} \right)}\, dx$$$

解答

$$$u=\frac{\pi m x}{a}$$$

$$$du=\left(\frac{\pi m x}{a}\right)^{\prime }dx = \frac{\pi m}{a} dx$$$ (步驟見»),並可得 $$$dx = \frac{a du}{\pi m}$$$

該積分變為

$${\color{red}{\int{\sin^{2}{\left(\frac{\pi m x}{a} \right)} d x}}} = {\color{red}{\int{\frac{a \sin^{2}{\left(u \right)}}{\pi m} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{a}{\pi m}$$$$$$f{\left(u \right)} = \sin^{2}{\left(u \right)}$$$

$${\color{red}{\int{\frac{a \sin^{2}{\left(u \right)}}{\pi m} d u}}} = {\color{red}{\frac{a \int{\sin^{2}{\left(u \right)} d u}}{\pi m}}}$$

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

$$\frac{a {\color{red}{\int{\sin^{2}{\left(u \right)} d u}}}}{\pi m} = \frac{a {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{\pi m}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(u \right)} = 1 - \cos{\left(2 u \right)}$$$

$$\frac{a {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{\pi m} = \frac{a {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}{2}\right)}}}{\pi m}$$

逐項積分:

$$\frac{a {\color{red}{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}}}{2 \pi m} = \frac{a {\color{red}{\left(\int{1 d u} - \int{\cos{\left(2 u \right)} d u}\right)}}}{2 \pi m}$$

配合 $$$c=1$$$,應用常數法則 $$$\int c\, du = c u$$$

$$\frac{a \left(- \int{\cos{\left(2 u \right)} d u} + {\color{red}{\int{1 d u}}}\right)}{2 \pi m} = \frac{a \left(- \int{\cos{\left(2 u \right)} d u} + {\color{red}{u}}\right)}{2 \pi m}$$

$$$v=2 u$$$

$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步驟見»),並可得 $$$du = \frac{dv}{2}$$$

該積分變為

$$\frac{a \left(u - {\color{red}{\int{\cos{\left(2 u \right)} d u}}}\right)}{2 \pi m} = \frac{a \left(u - {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)}{2 \pi m}$$

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(v \right)} = \cos{\left(v \right)}$$$

$$\frac{a \left(u - {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)}{2 \pi m} = \frac{a \left(u - {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}\right)}{2 \pi m}$$

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

$$\frac{a \left(u - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{2}\right)}{2 \pi m} = \frac{a \left(u - \frac{{\color{red}{\sin{\left(v \right)}}}}{2}\right)}{2 \pi m}$$

回顧一下 $$$v=2 u$$$

$$\frac{a \left(u - \frac{\sin{\left({\color{red}{v}} \right)}}{2}\right)}{2 \pi m} = \frac{a \left(u - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{2}\right)}{2 \pi m}$$

回顧一下 $$$u=\frac{\pi m x}{a}$$$

$$\frac{a \left(- \frac{\sin{\left(2 {\color{red}{u}} \right)}}{2} + {\color{red}{u}}\right)}{2 \pi m} = \frac{a \left(- \frac{\sin{\left(2 {\color{red}{\frac{\pi m x}{a}}} \right)}}{2} + {\color{red}{\frac{\pi m x}{a}}}\right)}{2 \pi m}$$

因此,

$$\int{\sin^{2}{\left(\frac{\pi m x}{a} \right)} d x} = \frac{a \left(- \frac{\sin{\left(\frac{2 \pi m x}{a} \right)}}{2} + \frac{\pi m x}{a}\right)}{2 \pi m}$$

化簡:

$$\int{\sin^{2}{\left(\frac{\pi m x}{a} \right)} d x} = - \frac{a \sin{\left(\frac{2 \pi m x}{a} \right)}}{4 \pi m} + \frac{x}{2}$$

加上積分常數:

$$\int{\sin^{2}{\left(\frac{\pi m x}{a} \right)} d x} = - \frac{a \sin{\left(\frac{2 \pi m x}{a} \right)}}{4 \pi m} + \frac{x}{2}+C$$

答案

$$$\int \sin^{2}{\left(\frac{\pi m x}{a} \right)}\, dx = \left(- \frac{a \sin{\left(\frac{2 \pi m x}{a} \right)}}{4 \pi m} + \frac{x}{2}\right) + C$$$A