$$$\sin^{4}{\left(6 x \right)}$$$ 的积分
您的输入
求$$$\int \sin^{4}{\left(6 x \right)}\, dx$$$。
解答
设$$$u=6 x$$$。
则$$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (步骤见»),并有$$$dx = \frac{du}{6}$$$。
因此,
$${\color{red}{\int{\sin^{4}{\left(6 x \right)} d x}}} = {\color{red}{\int{\frac{\sin^{4}{\left(u \right)}}{6} d u}}}$$
对 $$$c=\frac{1}{6}$$$ 和 $$$f{\left(u \right)} = \sin^{4}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\frac{\sin^{4}{\left(u \right)}}{6} d u}}} = {\color{red}{\left(\frac{\int{\sin^{4}{\left(u \right)} d u}}{6}\right)}}$$
应用降幂公式 $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$,并令 $$$\alpha= u $$$:
$$\frac{{\color{red}{\int{\sin^{4}{\left(u \right)} d u}}}}{6} = \frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{6}$$
对 $$$c=\frac{1}{8}$$$ 和 $$$f{\left(u \right)} = - 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$\frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{6} = \frac{{\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}{8}\right)}}}{6}$$
逐项积分:
$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}}}{48} = \frac{{\color{red}{\left(\int{3 d u} - \int{4 \cos{\left(2 u \right)} d u} + \int{\cos{\left(4 u \right)} d u}\right)}}}{48}$$
应用常数法则 $$$\int c\, du = c u$$$,使用 $$$c=3$$$:
$$- \frac{\int{4 \cos{\left(2 u \right)} d u}}{48} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} + \frac{{\color{red}{\int{3 d u}}}}{48} = - \frac{\int{4 \cos{\left(2 u \right)} d u}}{48} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} + \frac{{\color{red}{\left(3 u\right)}}}{48}$$
对 $$$c=4$$$ 和 $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{4 \cos{\left(2 u \right)} d u}}}}{48} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 u \right)} d u}\right)}}}{48}$$
设$$$v=2 u$$$。
则$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»),并有$$$du = \frac{dv}{2}$$$。
因此,
$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{12} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{12}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{12} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{12}$$
余弦函数的积分为 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:
$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{24} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\sin{\left(v \right)}}}}{24}$$
回忆一下 $$$v=2 u$$$:
$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{\sin{\left({\color{red}{v}} \right)}}{24} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{24}$$
设$$$v=4 u$$$。
则$$$dv=\left(4 u\right)^{\prime }du = 4 du$$$ (步骤见»),并有$$$du = \frac{dv}{4}$$$。
积分变为
$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{48} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{48}$$
对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{48} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{48}$$
余弦函数的积分为 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:
$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{192} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\sin{\left(v \right)}}}}{192}$$
回忆一下 $$$v=4 u$$$:
$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{\sin{\left({\color{red}{v}} \right)}}{192} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{192}$$
回忆一下 $$$u=6 x$$$:
$$- \frac{\sin{\left(2 {\color{red}{u}} \right)}}{24} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{192} + \frac{{\color{red}{u}}}{16} = - \frac{\sin{\left(2 {\color{red}{\left(6 x\right)}} \right)}}{24} + \frac{\sin{\left(4 {\color{red}{\left(6 x\right)}} \right)}}{192} + \frac{{\color{red}{\left(6 x\right)}}}{16}$$
因此,
$$\int{\sin^{4}{\left(6 x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(12 x \right)}}{24} + \frac{\sin{\left(24 x \right)}}{192}$$
化简:
$$\int{\sin^{4}{\left(6 x \right)} d x} = \frac{72 x - 8 \sin{\left(12 x \right)} + \sin{\left(24 x \right)}}{192}$$
加上积分常数:
$$\int{\sin^{4}{\left(6 x \right)} d x} = \frac{72 x - 8 \sin{\left(12 x \right)} + \sin{\left(24 x \right)}}{192}+C$$
答案
$$$\int \sin^{4}{\left(6 x \right)}\, dx = \frac{72 x - 8 \sin{\left(12 x \right)} + \sin{\left(24 x \right)}}{192} + C$$$A