$$$7 \tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$の積分

$$$\int 7 \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx$$$ 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、$$$c=7$$$ と $$$f{\left(x \right)} = \tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$ に対して適用する:

$${\color{red}{\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\left(7 \int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}\right)}}$$

正接を1つ取り出し、公式 $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$ を用いて、残りはすべて正割で表せ:

$$7 {\color{red}{\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = 7 {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}}$$

$$$u=\sec{\left(x \right)}$$$ とする。

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

したがって、

$$7 {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}} = 7 {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$

項別に積分せよ:

$$7 {\color{red}{\int{\left(u^{2} - 1\right)d u}}} = 7 {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u}\right)}}$$

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

$$7 \int{u^{2} d u} - 7 {\color{red}{\int{1 d u}}} = 7 \int{u^{2} d u} - 7 {\color{red}{u}}$$

$$$n=2$$$ 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:

$$- 7 u + 7 {\color{red}{\int{u^{2} d u}}}=- 7 u + 7 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 7 u + 7 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$- 7 {\color{red}{u}} + \frac{7 {\color{red}{u}}^{3}}{3} = - 7 {\color{red}{\sec{\left(x \right)}}} + \frac{7 {\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$

したがって、

$$\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{7 \sec^{3}{\left(x \right)}}{3} - 7 \sec{\left(x \right)}$$

簡単化せよ:

$$\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{7 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3}$$

積分定数を加える:

$$\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{7 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3}+C$$

$$$\int 7 \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx = \frac{7 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3} + C$$$A