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$$$\sin^{6}{\left(x \right)}$$$の積分
入力内容
$$$\int \sin^{6}{\left(x \right)}\, dx$$$ を求めよ。
解答
冪低減公式 $$$\sin^{6}{\left(\alpha \right)} = - \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} - \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$$$ を $$$\alpha=x$$$ に適用する:
$${\color{red}{\int{\sin^{6}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \frac{15 \cos{\left(2 x \right)}}{32} + \frac{3 \cos{\left(4 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{32} + \frac{5}{16}\right)d x}}}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{32}$$$ と $$$f{\left(x \right)} = - 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} - \cos{\left(6 x \right)} + 10$$$ に対して適用する:
$${\color{red}{\int{\left(- \frac{15 \cos{\left(2 x \right)}}{32} + \frac{3 \cos{\left(4 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{32} + \frac{5}{16}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(- 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} - \cos{\left(6 x \right)} + 10\right)d x}}{32}\right)}}$$
項別に積分せよ:
$$\frac{{\color{red}{\int{\left(- 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} - \cos{\left(6 x \right)} + 10\right)d x}}}}{32} = \frac{{\color{red}{\left(\int{10 d x} - \int{15 \cos{\left(2 x \right)} d x} + \int{6 \cos{\left(4 x \right)} d x} - \int{\cos{\left(6 x \right)} d x}\right)}}}{32}$$
$$$c=10$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:
$$- \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{{\color{red}{\int{10 d x}}}}{32} = - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{{\color{red}{\left(10 x\right)}}}{32}$$
$$$u=6 x$$$ とする。
すると $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{6}$$$ となります。
したがって、
$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{32}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{6}$$$ と $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ に対して適用する:
$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{32} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{32}$$
余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{192} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\sin{\left(u \right)}}}}{192}$$
次のことを思い出してください $$$u=6 x$$$:
$$\frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\sin{\left({\color{red}{u}} \right)}}{192} = \frac{5 x}{16} - \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{192}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=15$$$ と $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$ に対して適用する:
$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\int{15 \cos{\left(2 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{{\color{red}{\left(15 \int{\cos{\left(2 x \right)} d x}\right)}}}{32}$$
$$$u=2 x$$$ とする。
すると $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{2}$$$ となります。
積分は次のようになります
$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{32}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ に対して適用する:
$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{32} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{32}$$
余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 {\color{red}{\sin{\left(u \right)}}}}{64}$$
次のことを思い出してください $$$u=2 x$$$:
$$\frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{5 x}{16} - \frac{\sin{\left(6 x \right)}}{192} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{32} - \frac{15 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{64}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=6$$$ と $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$ に対して適用する:
$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{{\color{red}{\int{6 \cos{\left(4 x \right)} d x}}}}{32} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{{\color{red}{\left(6 \int{\cos{\left(4 x \right)} d x}\right)}}}{32}$$
$$$u=4 x$$$ とする。
すると $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{4}$$$ となります。
したがって、
$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{16} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{4}$$$ と $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ に対して適用する:
$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{16} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{16}$$
余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{64}$$
次のことを思い出してください $$$u=4 x$$$:
$$\frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192} + \frac{3 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{64}$$
したがって、
$$\int{\sin^{6}{\left(x \right)} d x} = \frac{5 x}{16} - \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} - \frac{\sin{\left(6 x \right)}}{192}$$
簡単化せよ:
$$\int{\sin^{6}{\left(x \right)} d x} = \frac{60 x - 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} - \sin{\left(6 x \right)}}{192}$$
積分定数を加える:
$$\int{\sin^{6}{\left(x \right)} d x} = \frac{60 x - 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} - \sin{\left(6 x \right)}}{192}+C$$
解答
$$$\int \sin^{6}{\left(x \right)}\, dx = \frac{60 x - 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} - \sin{\left(6 x \right)}}{192} + C$$$A