$$$\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}$$$ 的積分

求$$$\int \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx$$$。

先提出一個餘弦，並將其餘部分以正弦表示，使用公式 $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$，令 $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{2}{\left(x \right)} \cos{\left(x \right)} d x}}}$$

令 $$$u=\sin{\left(x \right)}$$$。

則 $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (步驟見»)，並可得 $$$\cos{\left(x \right)} dx = du$$$。

所以，

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

Expand the expression:

$${\color{red}{\int{u^{2} \left(1 - u^{2}\right)^{2} d u}}} = {\color{red}{\int{\left(u^{6} - 2 u^{4} + u^{2}\right)d u}}}$$

逐項積分：

$${\color{red}{\int{\left(u^{6} - 2 u^{4} + u^{2}\right)d u}}} = {\color{red}{\left(\int{u^{2} d u} - \int{2 u^{4} d u} + \int{u^{6} d u}\right)}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

$$- \int{2 u^{4} d u} + \int{u^{6} d u} + {\color{red}{\int{u^{2} d u}}}=- \int{2 u^{4} d u} + \int{u^{6} d u} + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- \int{2 u^{4} d u} + \int{u^{6} d u} + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=6$$$：

$$\frac{u^{3}}{3} - \int{2 u^{4} d u} + {\color{red}{\int{u^{6} d u}}}=\frac{u^{3}}{3} - \int{2 u^{4} d u} + {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=\frac{u^{3}}{3} - \int{2 u^{4} d u} + {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=2$$$ 與 $$$f{\left(u \right)} = u^{4}$$$：

$$\frac{u^{7}}{7} + \frac{u^{3}}{3} - {\color{red}{\int{2 u^{4} d u}}} = \frac{u^{7}}{7} + \frac{u^{3}}{3} - {\color{red}{\left(2 \int{u^{4} d u}\right)}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=4$$$：

$$\frac{u^{7}}{7} + \frac{u^{3}}{3} - 2 {\color{red}{\int{u^{4} d u}}}=\frac{u^{7}}{7} + \frac{u^{3}}{3} - 2 {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\frac{u^{7}}{7} + \frac{u^{3}}{3} - 2 {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

回顧一下 $$$u=\sin{\left(x \right)}$$$：

$$\frac{{\color{red}{u}}^{3}}{3} - \frac{2 {\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{7}}{7} = \frac{{\color{red}{\sin{\left(x \right)}}}^{3}}{3} - \frac{2 {\color{red}{\sin{\left(x \right)}}}^{5}}{5} + \frac{{\color{red}{\sin{\left(x \right)}}}^{7}}{7}$$

因此，

$$\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}$$

化簡：

$$\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}$$

加上積分常數：

$$\int{\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}+C$$

$$$\int \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx = \frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105} + C$$$A