$$$\left(2 \sin{\left(\theta \right)} + 3\right)^{2}$$$の積分

$$$\int \left(2 \sin{\left(\theta \right)} + 3\right)^{2}\, d\theta$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{\left(2 \sin{\left(\theta \right)} + 3\right)^{2} d \theta}}} = {\color{red}{\int{\left(4 \sin^{2}{\left(\theta \right)} + 12 \sin{\left(\theta \right)} + 9\right)d \theta}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(4 \sin^{2}{\left(\theta \right)} + 12 \sin{\left(\theta \right)} + 9\right)d \theta}}} = {\color{red}{\left(\int{9 d \theta} + \int{12 \sin{\left(\theta \right)} d \theta} + \int{4 \sin^{2}{\left(\theta \right)} d \theta}\right)}}$$

$$$c=9$$$ に対して定数則 $$$\int c\, d\theta = c \theta$$$ を適用する:

$$\int{12 \sin{\left(\theta \right)} d \theta} + \int{4 \sin^{2}{\left(\theta \right)} d \theta} + {\color{red}{\int{9 d \theta}}} = \int{12 \sin{\left(\theta \right)} d \theta} + \int{4 \sin^{2}{\left(\theta \right)} d \theta} + {\color{red}{\left(9 \theta\right)}}$$

定数倍の法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ を、$$$c=4$$$ と $$$f{\left(\theta \right)} = \sin^{2}{\left(\theta \right)}$$$ に対して適用する:

$$9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + {\color{red}{\int{4 \sin^{2}{\left(\theta \right)} d \theta}}} = 9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + {\color{red}{\left(4 \int{\sin^{2}{\left(\theta \right)} d \theta}\right)}}$$

冪低減公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ を $$$\alpha=\theta$$$ に適用する:

$$9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + 4 {\color{red}{\int{\sin^{2}{\left(\theta \right)} d \theta}}} = 9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + 4 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 \theta \right)}}{2}\right)d \theta}}}$$

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

$$9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + 4 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 \theta \right)}}{2}\right)d \theta}}} = 9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + 4 {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 \theta \right)}\right)d \theta}}{2}\right)}}$$

項別に積分せよ:

$$9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + 2 {\color{red}{\int{\left(1 - \cos{\left(2 \theta \right)}\right)d \theta}}} = 9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} + 2 {\color{red}{\left(\int{1 d \theta} - \int{\cos{\left(2 \theta \right)} d \theta}\right)}}$$

$$$c=1$$$ に対して定数則 $$$\int c\, d\theta = c \theta$$$ を適用する:

$$9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - 2 \int{\cos{\left(2 \theta \right)} d \theta} + 2 {\color{red}{\int{1 d \theta}}} = 9 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - 2 \int{\cos{\left(2 \theta \right)} d \theta} + 2 {\color{red}{\theta}}$$

$$$u=2 \theta$$$ とする。

すると $$$du=\left(2 \theta\right)^{\prime }d\theta = 2 d\theta$$$（手順は»で確認できます）、$$$d\theta = \frac{du}{2}$$$ となります。

したがって、

$$11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - 2 {\color{red}{\int{\cos{\left(2 \theta \right)} d \theta}}} = 11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

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:

$$11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = 11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - 2 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - {\color{red}{\int{\cos{\left(u \right)} d u}}} = 11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - {\color{red}{\sin{\left(u \right)}}}$$

次のことを思い出してください $$$u=2 \theta$$$:

$$11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - \sin{\left({\color{red}{u}} \right)} = 11 \theta + \int{12 \sin{\left(\theta \right)} d \theta} - \sin{\left({\color{red}{\left(2 \theta\right)}} \right)}$$

定数倍の法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ を、$$$c=12$$$ と $$$f{\left(\theta \right)} = \sin{\left(\theta \right)}$$$ に対して適用する:

$$11 \theta - \sin{\left(2 \theta \right)} + {\color{red}{\int{12 \sin{\left(\theta \right)} d \theta}}} = 11 \theta - \sin{\left(2 \theta \right)} + {\color{red}{\left(12 \int{\sin{\left(\theta \right)} d \theta}\right)}}$$

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

$$11 \theta - \sin{\left(2 \theta \right)} + 12 {\color{red}{\int{\sin{\left(\theta \right)} d \theta}}} = 11 \theta - \sin{\left(2 \theta \right)} + 12 {\color{red}{\left(- \cos{\left(\theta \right)}\right)}}$$

したがって、

$$\int{\left(2 \sin{\left(\theta \right)} + 3\right)^{2} d \theta} = 11 \theta - \sin{\left(2 \theta \right)} - 12 \cos{\left(\theta \right)}$$

積分定数を加える:

$$\int{\left(2 \sin{\left(\theta \right)} + 3\right)^{2} d \theta} = 11 \theta - \sin{\left(2 \theta \right)} - 12 \cos{\left(\theta \right)}+C$$

$$$\int \left(2 \sin{\left(\theta \right)} + 3\right)^{2}\, d\theta = \left(11 \theta - \sin{\left(2 \theta \right)} - 12 \cos{\left(\theta \right)}\right) + C$$$A