$$$\sin^{3}{\left(x \right)} \cos{\left(2 x \right)}$$$の積分

$$$\int \sin^{3}{\left(x \right)} \cos{\left(2 x \right)}\, dx$$$ を求めよ。

冪低減公式 $$$\sin^{3}{\left(\alpha \right)} = \frac{3 \sin{\left(\alpha \right)}}{4} - \frac{\sin{\left(3 \alpha \right)}}{4}$$$ を $$$\alpha=x$$$ に適用する:

$${\color{red}{\int{\sin^{3}{\left(x \right)} \cos{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \cos{\left(2 x \right)}}{4} d x}}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{4}$$$ と $$$f{\left(x \right)} = \left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \cos{\left(2 x \right)}$$$ に対して適用する:

$${\color{red}{\int{\frac{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \cos{\left(2 x \right)}}{4} d x}}} = {\color{red}{\left(\frac{\int{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \cos{\left(2 x \right)} d x}}{4}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \cos{\left(2 x \right)} d x}}}}{4} = \frac{{\color{red}{\int{\left(3 \sin{\left(x \right)} \cos{\left(2 x \right)} - \sin{\left(3 x \right)} \cos{\left(2 x \right)}\right)d x}}}}{4}$$

項別に積分せよ:

$$\frac{{\color{red}{\int{\left(3 \sin{\left(x \right)} \cos{\left(2 x \right)} - \sin{\left(3 x \right)} \cos{\left(2 x \right)}\right)d x}}}}{4} = \frac{{\color{red}{\left(\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x} - \int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}\right)}}}{4}$$

公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ を用い、$$$\alpha=3 x$$$ と $$$\beta=2 x$$$ を使って被積分関数を書き換えなさい。:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}}}{4} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{4}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(x \right)} = \sin{\left(x \right)} + \sin{\left(5 x \right)}$$$ に対して適用する:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{4} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}{2}\right)}}}{4}$$

項別に積分せよ:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}}}{8} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(\int{\sin{\left(x \right)} d x} + \int{\sin{\left(5 x \right)} d x}\right)}}}{8}$$

正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{\int{\sin{\left(5 x \right)} d x}}{8} - \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{8} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{\int{\sin{\left(5 x \right)} d x}}{8} - \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{8}$$

$$$u=5 x$$$ とする。

すると $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{5}$$$ となります。

したがって、

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{8}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{5}$$$ と $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ に対して適用する:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{8}$$

正弦関数の不定積分は$$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$です:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{40} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{40}$$

次のことを思い出してください $$$u=5 x$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} + \frac{\cos{\left({\color{red}{u}} \right)}}{40} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} + \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{40}$$

$$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ の公式を用い、$$$\alpha=x$$$ および $$$\beta=2 x$$$ を用いて $$$\sin\left(x \right)\cos\left(2 x \right)$$$ を変形せよ:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}}}{4} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{2} + \frac{3 \sin{\left(3 x \right)}}{2}\right)d x}}}}{4}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(x \right)} = - 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}$$$ に対して適用する:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{2} + \frac{3 \sin{\left(3 x \right)}}{2}\right)d x}}}}{4} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(\frac{\int{\left(- 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}\right)d x}}{2}\right)}}}{4}$$

項別に積分せよ:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\left(- 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}\right)d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(- \int{3 \sin{\left(x \right)} d x} + \int{3 \sin{\left(3 x \right)} d x}\right)}}}{8}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=3$$$ と $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ に対して適用する:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{3 \sin{\left(x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(3 \int{\sin{\left(x \right)} d x}\right)}}}{8}$$

正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\int{\sin{\left(x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\left(- \cos{\left(x \right)}\right)}}}{8}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=3$$$ と $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$ に対して適用する:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{3 \sin{\left(3 x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(3 \int{\sin{\left(3 x \right)} d x}\right)}}}{8}$$

$$$u=3 x$$$ とする。

すると $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{3}$$$ となります。

したがって、

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\int{\sin{\left(3 x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{8}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{3}$$$ と $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ に対して適用する:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}}{8}$$

正弦関数の不定積分は$$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$です:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

次のことを思い出してください $$$u=3 x$$$:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} - \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{8}$$

したがって、

$$\int{\sin^{3}{\left(x \right)} \cos{\left(2 x \right)} d x} = \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40}$$

積分定数を加える:

$$\int{\sin^{3}{\left(x \right)} \cos{\left(2 x \right)} d x} = \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40}+C$$

$$$\int \sin^{3}{\left(x \right)} \cos{\left(2 x \right)}\, dx = \left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40}\right) + C$$$A