$$$\sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)}$$$ 的積分
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您的輸入
求$$$\int \sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$。
解答
提出一個正弦因子,將其餘部分用餘弦表示,使用公式 $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$,其中 $$$\alpha=x$$$:
$${\color{red}{\int{\sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{3}{\left(x \right)} d x}}}$$
令 $$$u=\cos{\left(x \right)}$$$。
則 $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (步驟見»),並可得 $$$\sin{\left(x \right)} dx = - du$$$。
因此,
$${\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{3} \left(1 - u^{2}\right)\right)d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = u^{3} \left(1 - u^{2}\right)$$$:
$${\color{red}{\int{\left(- u^{3} \left(1 - u^{2}\right)\right)d u}}} = {\color{red}{\left(- \int{u^{3} \left(1 - u^{2}\right) d u}\right)}}$$
Expand the expression:
$$- {\color{red}{\int{u^{3} \left(1 - u^{2}\right) d u}}} = - {\color{red}{\int{\left(- u^{5} + u^{3}\right)d u}}}$$
逐項積分:
$$- {\color{red}{\int{\left(- u^{5} + u^{3}\right)d u}}} = - {\color{red}{\left(\int{u^{3} d u} - \int{u^{5} d u}\right)}}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=3$$$:
$$\int{u^{5} d u} - {\color{red}{\int{u^{3} d u}}}=\int{u^{5} d u} - {\color{red}{\frac{u^{1 + 3}}{1 + 3}}}=\int{u^{5} d u} - {\color{red}{\left(\frac{u^{4}}{4}\right)}}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=5$$$:
$$- \frac{u^{4}}{4} + {\color{red}{\int{u^{5} d u}}}=- \frac{u^{4}}{4} + {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=- \frac{u^{4}}{4} + {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$
回顧一下 $$$u=\cos{\left(x \right)}$$$:
$$- \frac{{\color{red}{u}}^{4}}{4} + \frac{{\color{red}{u}}^{6}}{6} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{4}}{4} + \frac{{\color{red}{\cos{\left(x \right)}}}^{6}}{6}$$
因此,
$$\int{\sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\cos^{6}{\left(x \right)}}{6} - \frac{\cos^{4}{\left(x \right)}}{4}$$
加上積分常數:
$$\int{\sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\cos^{6}{\left(x \right)}}{6} - \frac{\cos^{4}{\left(x \right)}}{4}+C$$
答案
$$$\int \sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(\frac{\cos^{6}{\left(x \right)}}{6} - \frac{\cos^{4}{\left(x \right)}}{4}\right) + C$$$A