$$$\left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)}$$$ 的積分
相關計算器: 定積分與廣義積分計算器
您的輸入
求$$$\int \left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)}\, dx$$$。
解答
Expand the expression:
$${\color{red}{\int{\left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x}}} = {\color{red}{\int{\left(- 3 \sin^{3}{\left(x \right)} + 2 \sin{\left(x \right)}\right)d x}}}$$
逐項積分:
$${\color{red}{\int{\left(- 3 \sin^{3}{\left(x \right)} + 2 \sin{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{2 \sin{\left(x \right)} d x} - \int{3 \sin^{3}{\left(x \right)} d x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=3$$$ 與 $$$f{\left(x \right)} = \sin^{3}{\left(x \right)}$$$:
$$\int{2 \sin{\left(x \right)} d x} - {\color{red}{\int{3 \sin^{3}{\left(x \right)} d x}}} = \int{2 \sin{\left(x \right)} d x} - {\color{red}{\left(3 \int{\sin^{3}{\left(x \right)} d x}\right)}}$$
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 $$$\alpha=x$$$:
$$\int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\int{\sin^{3}{\left(x \right)} d x}}} = \int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x}}}$$
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,
$$\int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x}}} = \int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$
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 $$$f{\left(u \right)} = 1 - u^{2}$$$:
$$\int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\int{\left(u^{2} - 1\right)d u}}} = \int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\left(- \int{\left(1 - u^{2}\right)d u}\right)}}$$
逐項積分:
$$\int{2 \sin{\left(x \right)} d x} + 3 {\color{red}{\int{\left(1 - u^{2}\right)d u}}} = \int{2 \sin{\left(x \right)} d x} + 3 {\color{red}{\left(\int{1 d u} - \int{u^{2} d u}\right)}}$$
配合 $$$c=1$$$,應用常數法則 $$$\int c\, du = c u$$$:
$$\int{2 \sin{\left(x \right)} d x} - 3 \int{u^{2} d u} + 3 {\color{red}{\int{1 d u}}} = \int{2 \sin{\left(x \right)} d x} - 3 \int{u^{2} d u} + 3 {\color{red}{u}}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=2$$$:
$$3 u + \int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\int{u^{2} d u}}}=3 u + \int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=3 u + \int{2 \sin{\left(x \right)} d x} - 3 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
回顧一下 $$$u=\cos{\left(x \right)}$$$:
$$\int{2 \sin{\left(x \right)} d x} + 3 {\color{red}{u}} - {\color{red}{u}}^{3} = \int{2 \sin{\left(x \right)} d x} + 3 {\color{red}{\cos{\left(x \right)}}} - {\color{red}{\cos{\left(x \right)}}}^{3}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = \sin{\left(x \right)}$$$:
$$- \cos^{3}{\left(x \right)} + 3 \cos{\left(x \right)} + {\color{red}{\int{2 \sin{\left(x \right)} d x}}} = - \cos^{3}{\left(x \right)} + 3 \cos{\left(x \right)} + {\color{red}{\left(2 \int{\sin{\left(x \right)} d x}\right)}}$$
正弦函數的積分為 $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:
$$- \cos^{3}{\left(x \right)} + 3 \cos{\left(x \right)} + 2 {\color{red}{\int{\sin{\left(x \right)} d x}}} = - \cos^{3}{\left(x \right)} + 3 \cos{\left(x \right)} + 2 {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$
因此,
$$\int{\left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x} = - \cos^{3}{\left(x \right)} + \cos{\left(x \right)}$$
加上積分常數:
$$\int{\left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x} = - \cos^{3}{\left(x \right)} + \cos{\left(x \right)}+C$$
答案
$$$\int \left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)}\, dx = \left(- \cos^{3}{\left(x \right)} + \cos{\left(x \right)}\right) + C$$$A