$$$\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}$$$ 的積分

此計算器將求出 $$$\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}$$$ 的不定積分(原函數),並顯示步驟。

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您的輸入

$$$\int \sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}\, d\theta$$$

解答

使用倍角公式 $$$\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1}{2}\sin\left( 2 \theta \right)$$$ 改寫被積函數:

$${\color{red}{\int{\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\sin^{2}{\left(2 \theta \right)}}{4} d \theta}}}$$

套用常數倍法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$,使用 $$$c=\frac{1}{4}$$$$$$f{\left(\theta \right)} = \sin^{2}{\left(2 \theta \right)}$$$

$${\color{red}{\int{\frac{\sin^{2}{\left(2 \theta \right)}}{4} d \theta}}} = {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 \theta \right)} d \theta}}{4}\right)}}$$

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

$$\frac{{\color{red}{\int{\sin^{2}{\left(2 \theta \right)} d \theta}}}}{4} = \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 \theta \right)}}{2}\right)d \theta}}}}{4}$$

套用常數倍法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$,使用 $$$c=\frac{1}{2}$$$$$$f{\left(\theta \right)} = 1 - \cos{\left(4 \theta \right)}$$$

$$\frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 \theta \right)}}{2}\right)d \theta}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(4 \theta \right)}\right)d \theta}}{2}\right)}}}{4}$$

逐項積分:

$$\frac{{\color{red}{\int{\left(1 - \cos{\left(4 \theta \right)}\right)d \theta}}}}{8} = \frac{{\color{red}{\left(\int{1 d \theta} - \int{\cos{\left(4 \theta \right)} d \theta}\right)}}}{8}$$

配合 $$$c=1$$$,應用常數法則 $$$\int c\, d\theta = c \theta$$$

$$- \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{1 d \theta}}}}{8} = - \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\theta}}}{8}$$

$$$u=4 \theta$$$

$$$du=\left(4 \theta\right)^{\prime }d\theta = 4 d\theta$$$ (步驟見»),並可得 $$$d\theta = \frac{du}{4}$$$

因此,

$$\frac{\theta}{8} - \frac{{\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}}}{8} = \frac{\theta}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{4}$$$$$$f{\left(u \right)} = \cos{\left(u \right)}$$$

$$\frac{\theta}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{\theta}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

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

$$\frac{\theta}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{\theta}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

回顧一下 $$$u=4 \theta$$$

$$\frac{\theta}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{\theta}{8} - \frac{\sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{32}$$

因此,

$$\int{\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)} d \theta} = \frac{\theta}{8} - \frac{\sin{\left(4 \theta \right)}}{32}$$

加上積分常數:

$$\int{\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)} d \theta} = \frac{\theta}{8} - \frac{\sin{\left(4 \theta \right)}}{32}+C$$

答案

$$$\int \sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}\, d\theta = \left(\frac{\theta}{8} - \frac{\sin{\left(4 \theta \right)}}{32}\right) + C$$$A


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