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

$$$\int \tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)}\, dx$$$ を求めよ。

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:

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

$$$u=\sec{\left(x \right)}$$$ 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。

積分は次のようになります

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

Expand the expression:

$${\color{red}{\int{u^{2} \left(u^{2} - 1\right) d u}}} = {\color{red}{\int{\left(u^{4} - u^{2}\right)d u}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(u^{4} - u^{2}\right)d u}}} = {\color{red}{\left(- \int{u^{2} d u} + \int{u^{4} d u}\right)}}$$

$$$n=4$$$ 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:

$$- \int{u^{2} d u} + {\color{red}{\int{u^{4} d u}}}=- \int{u^{2} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=- \int{u^{2} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

$$$n=2$$$ 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:

$$\frac{u^{5}}{5} - {\color{red}{\int{u^{2} d u}}}=\frac{u^{5}}{5} - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\frac{u^{5}}{5} - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$- \frac{{\color{red}{u}}^{3}}{3} + \frac{{\color{red}{u}}^{5}}{5} = - \frac{{\color{red}{\sec{\left(x \right)}}}^{3}}{3} + \frac{{\color{red}{\sec{\left(x \right)}}}^{5}}{5}$$

したがって、

$$\int{\tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}$$

積分定数を加える:

$$\int{\tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}+C$$

$$$\int \tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)}\, dx = \left(\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}\right) + C$$$A