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

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

Expand the expression:

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

項別に積分せよ:

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

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

$$- \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + {\color{red}{\int{\frac{2}{\cos^{2}{\left(x \right)}} d x}}} = - \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + {\color{red}{\left(2 \int{\frac{1}{\cos^{2}{\left(x \right)}} d x}\right)}}$$

被積分関数を正割関数で表しなさい:

$$- \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + 2 {\color{red}{\int{\frac{1}{\cos^{2}{\left(x \right)}} d x}}} = - \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + 2 {\color{red}{\int{\sec^{2}{\left(x \right)} d x}}}$$

$$$\sec^{2}{\left(x \right)}$$$ の不定積分は $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$ です:

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

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

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

$$$u=\cos{\left(x \right)}$$$ とする。

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

したがって、

$$2 \tan{\left(x \right)} - 3 {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = 2 \tan{\left(x \right)} - 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}}$$

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

$$2 \tan{\left(x \right)} - 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}} = 2 \tan{\left(x \right)} - 3 {\color{red}{\left(- \int{\frac{1}{u^{2}} d u}\right)}}$$

$$$n=-2$$$ を用いて、べき乗の法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$2 \tan{\left(x \right)} + 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=2 \tan{\left(x \right)} + 3 {\color{red}{\int{u^{-2} d u}}}=2 \tan{\left(x \right)} + 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=2 \tan{\left(x \right)} + 3 {\color{red}{\left(- u^{-1}\right)}}=2 \tan{\left(x \right)} + 3 {\color{red}{\left(- \frac{1}{u}\right)}}$$

次のことを思い出してください $$$u=\cos{\left(x \right)}$$$:

$$2 \tan{\left(x \right)} - 3 {\color{red}{u}}^{-1} = 2 \tan{\left(x \right)} - 3 {\color{red}{\cos{\left(x \right)}}}^{-1}$$

したがって、

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

積分定数を加える:

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

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