$$$16 \cos{\left(x \right)} \tan{\left(x \right)}$$$の積分

$$$\int 16 \cos{\left(x \right)} \tan{\left(x \right)}\, dx$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=16$$$ と $$$f{\left(x \right)} = \cos{\left(x \right)} \tan{\left(x \right)}$$$ に対して適用する:

$${\color{red}{\int{16 \cos{\left(x \right)} \tan{\left(x \right)} d x}}} = {\color{red}{\left(16 \int{\cos{\left(x \right)} \tan{\left(x \right)} d x}\right)}}$$

被積分関数を簡単化する:

$$16 {\color{red}{\int{\cos{\left(x \right)} \tan{\left(x \right)} d x}}} = 16 {\color{red}{\int{\sin{\left(x \right)} d x}}}$$

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

$$16 {\color{red}{\int{\sin{\left(x \right)} d x}}} = 16 {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

したがって、

$$\int{16 \cos{\left(x \right)} \tan{\left(x \right)} d x} = - 16 \cos{\left(x \right)}$$

積分定数を加える:

$$\int{16 \cos{\left(x \right)} \tan{\left(x \right)} d x} = - 16 \cos{\left(x \right)}+C$$

$$$\int 16 \cos{\left(x \right)} \tan{\left(x \right)}\, dx = - 16 \cos{\left(x \right)} + C$$$A