$$$\frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}}$$$の積分

$$$\int \frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}}\, dx$$$ を求めよ。

分子と分母に$$$\frac{1}{\cos^{6}{\left(x \right)}}$$$を掛け、$$$\frac{\cos^{6}{\left(x \right)}}{\sin^{6}{\left(x \right)}}$$$を$$$\frac{1}{\tan^{6}{\left(x \right)}}$$$に変換します。:

$${\color{red}{\int{\frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cos^{12}{\left(x \right)} \tan^{6}{\left(x \right)}} d x}}}$$

余弦を2つくくり出し、公式 $$$\frac{1}{\cos^{2}{\left(x \right)}}=\sec^{2}{\left(x \right)}$$$ を用いて正割で表し直しなさい。:

$${\color{red}{\int{\frac{1}{\cos^{12}{\left(x \right)} \tan^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{10}{\left(x \right)} \tan^{6}{\left(x \right)}} d x}}}$$

公式 $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$ を用いて、余弦を正接で表せ。:

$${\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{10}{\left(x \right)} \tan^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\left(\tan^{2}{\left(x \right)} + 1\right)^{5} \sec^{2}{\left(x \right)}}{\tan^{6}{\left(x \right)}} d x}}}$$

$$$u=\tan{\left(x \right)}$$$ とする。

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

したがって、

$${\color{red}{\int{\frac{\left(\tan^{2}{\left(x \right)} + 1\right)^{5} \sec^{2}{\left(x \right)}}{\tan^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\left(u^{2} + 1\right)^{5}}{u^{6}} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{\left(u^{2} + 1\right)^{5}}{u^{6}} d u}}} = {\color{red}{\int{\left(u^{4} + 5 u^{2} + 10 + \frac{10}{u^{2}} + \frac{5}{u^{4}} + \frac{1}{u^{6}}\right)d u}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(u^{4} + 5 u^{2} + 10 + \frac{10}{u^{2}} + \frac{5}{u^{4}} + \frac{1}{u^{6}}\right)d u}}} = {\color{red}{\left(\int{10 d u} + \int{\frac{1}{u^{6}} d u} + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u}\right)}}$$

$$$c=10$$$ に対して定数則 $$$\int c\, du = c u$$$ を適用する:

$$\int{\frac{1}{u^{6}} d u} + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\int{10 d u}}} = \int{\frac{1}{u^{6}} d u} + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\left(10 u\right)}}$$

$$$n=-6$$$ 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:

$$10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\int{\frac{1}{u^{6}} d u}}}=10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\int{u^{-6} d u}}}=10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\frac{u^{-6 + 1}}{-6 + 1}}}=10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\left(- \frac{u^{-5}}{5}\right)}}=10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\left(- \frac{1}{5 u^{5}}\right)}}$$

$$$n=4$$$ 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:

$$10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + {\color{red}{\int{u^{4} d u}}} - \frac{1}{5 u^{5}}=10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}} - \frac{1}{5 u^{5}}=10 u + \int{\frac{5}{u^{4}} d u} + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}} - \frac{1}{5 u^{5}}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=5$$$ と $$$f{\left(u \right)} = \frac{1}{u^{4}}$$$ に対して適用する:

$$\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + {\color{red}{\int{\frac{5}{u^{4}} d u}}} - \frac{1}{5 u^{5}} = \frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + {\color{red}{\left(5 \int{\frac{1}{u^{4}} d u}\right)}} - \frac{1}{5 u^{5}}$$

$$$n=-4$$$ 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:

$$\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + 5 {\color{red}{\int{\frac{1}{u^{4}} d u}}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + 5 {\color{red}{\int{u^{-4} d u}}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + 5 {\color{red}{\frac{u^{-4 + 1}}{-4 + 1}}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + 5 {\color{red}{\left(- \frac{u^{-3}}{3}\right)}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + \int{5 u^{2} d u} + 5 {\color{red}{\left(- \frac{1}{3 u^{3}}\right)}} - \frac{1}{5 u^{5}}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=5$$$ と $$$f{\left(u \right)} = u^{2}$$$ に対して適用する:

$$\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + {\color{red}{\int{5 u^{2} d u}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}} = \frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + {\color{red}{\left(5 \int{u^{2} d u}\right)}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}$$

$$$n=2$$$ 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:

$$\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + 5 {\color{red}{\int{u^{2} d u}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + 5 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + 10 u + \int{\frac{10}{u^{2}} d u} + 5 {\color{red}{\left(\frac{u^{3}}{3}\right)}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=10$$$ と $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$ に対して適用する:

$$\frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + {\color{red}{\int{\frac{10}{u^{2}} d u}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}} = \frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + {\color{red}{\left(10 \int{\frac{1}{u^{2}} d u}\right)}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}$$

$$$n=-2$$$ 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:

$$\frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + 10 {\color{red}{\int{\frac{1}{u^{2}} d u}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + 10 {\color{red}{\int{u^{-2} d u}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + 10 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + 10 {\color{red}{\left(- u^{-1}\right)}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}=\frac{u^{5}}{5} + \frac{5 u^{3}}{3} + 10 u + 10 {\color{red}{\left(- \frac{1}{u}\right)}} - \frac{5}{3 u^{3}} - \frac{1}{5 u^{5}}$$

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

$$- \frac{{\color{red}{u}}^{-5}}{5} - \frac{5 {\color{red}{u}}^{-3}}{3} - 10 {\color{red}{u}}^{-1} + 10 {\color{red}{u}} + \frac{5 {\color{red}{u}}^{3}}{3} + \frac{{\color{red}{u}}^{5}}{5} = - \frac{{\color{red}{\tan{\left(x \right)}}}^{-5}}{5} - \frac{5 {\color{red}{\tan{\left(x \right)}}}^{-3}}{3} - 10 {\color{red}{\tan{\left(x \right)}}}^{-1} + 10 {\color{red}{\tan{\left(x \right)}}} + \frac{5 {\color{red}{\tan{\left(x \right)}}}^{3}}{3} + \frac{{\color{red}{\tan{\left(x \right)}}}^{5}}{5}$$

したがって、

$$\int{\frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5} + \frac{5 \tan^{3}{\left(x \right)}}{3} + 10 \tan{\left(x \right)} - \frac{10}{\tan{\left(x \right)}} - \frac{5}{3 \tan^{3}{\left(x \right)}} - \frac{1}{5 \tan^{5}{\left(x \right)}}$$

簡単化せよ:

$$\int{\frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}} d x} = \frac{\left(3 \tan^{4}{\left(x \right)} + 25 \tan^{2}{\left(x \right)} + 150\right) \tan^{6}{\left(x \right)} - 150 \tan^{4}{\left(x \right)} - 25 \tan^{2}{\left(x \right)} - 3}{15 \tan^{5}{\left(x \right)}}$$

積分定数を加える:

$$\int{\frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}} d x} = \frac{\left(3 \tan^{4}{\left(x \right)} + 25 \tan^{2}{\left(x \right)} + 150\right) \tan^{6}{\left(x \right)} - 150 \tan^{4}{\left(x \right)} - 25 \tan^{2}{\left(x \right)} - 3}{15 \tan^{5}{\left(x \right)}}+C$$

$$$\int \frac{1}{\sin^{6}{\left(x \right)} \cos^{6}{\left(x \right)}}\, dx = \frac{\left(3 \tan^{4}{\left(x \right)} + 25 \tan^{2}{\left(x \right)} + 150\right) \tan^{6}{\left(x \right)} - 150 \tan^{4}{\left(x \right)} - 25 \tan^{2}{\left(x \right)} - 3}{15 \tan^{5}{\left(x \right)}} + C$$$A