$$$\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}$$$の積分
関連する計算機: 定積分・広義積分計算機
入力内容
$$$\int \sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}\, d\theta$$$ を求めよ。
解答
二倍角の公式 $$$\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1}{2}\sin\left( 2 \theta \right)$$$ を用いて被積分関数を書き換えます:
$${\color{red}{\int{\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\sin^{2}{\left(2 \theta \right)}}{4} d \theta}}}$$
定数倍の法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ を、$$$c=\frac{1}{4}$$$ と $$$f{\left(\theta \right)} = \sin^{2}{\left(2 \theta \right)}$$$ に対して適用する:
$${\color{red}{\int{\frac{\sin^{2}{\left(2 \theta \right)}}{4} d \theta}}} = {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 \theta \right)} d \theta}}{4}\right)}}$$
冪低減公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ を $$$\alpha=2 \theta$$$ に適用する:
$$\frac{{\color{red}{\int{\sin^{2}{\left(2 \theta \right)} d \theta}}}}{4} = \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 \theta \right)}}{2}\right)d \theta}}}}{4}$$
定数倍の法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(\theta \right)} = 1 - \cos{\left(4 \theta \right)}$$$ に対して適用する:
$$\frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 \theta \right)}}{2}\right)d \theta}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(4 \theta \right)}\right)d \theta}}{2}\right)}}}{4}$$
項別に積分せよ:
$$\frac{{\color{red}{\int{\left(1 - \cos{\left(4 \theta \right)}\right)d \theta}}}}{8} = \frac{{\color{red}{\left(\int{1 d \theta} - \int{\cos{\left(4 \theta \right)} d \theta}\right)}}}{8}$$
$$$c=1$$$ に対して定数則 $$$\int c\, d\theta = c \theta$$$ を適用する:
$$- \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{1 d \theta}}}}{8} = - \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\theta}}}{8}$$
$$$u=4 \theta$$$ とする。
すると $$$du=\left(4 \theta\right)^{\prime }d\theta = 4 d\theta$$$(手順は»で確認できます)、$$$d\theta = \frac{du}{4}$$$ となります。
この積分は次のように書き換えられる
$$\frac{\theta}{8} - \frac{{\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}}}{8} = \frac{\theta}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{4}$$$ と $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ に対して適用する:
$$\frac{\theta}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{\theta}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$
余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{\theta}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{\theta}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$
次のことを思い出してください $$$u=4 \theta$$$:
$$\frac{\theta}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{\theta}{8} - \frac{\sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{32}$$
したがって、
$$\int{\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)} d \theta} = \frac{\theta}{8} - \frac{\sin{\left(4 \theta \right)}}{32}$$
積分定数を加える:
$$\int{\sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)} d \theta} = \frac{\theta}{8} - \frac{\sin{\left(4 \theta \right)}}{32}+C$$
解答
$$$\int \sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}\, d\theta = \left(\frac{\theta}{8} - \frac{\sin{\left(4 \theta \right)}}{32}\right) + C$$$A