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

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

項別に積分せよ:

$${\color{red}{\int{\left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x} + \int{\tan{\left(x \right)} d x} - \int{\cot{\left(x \right)} d x}\right)}}$$

余接関数を$$$\cot\left(x\right)=\frac{\cos\left(x\right)}{\sin\left(x\right)}$$$として書き換えなさい:

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

$$$u=\sin{\left(x \right)}$$$ とする。

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

したがって、

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

$$$\frac{1}{u}$$$ の不定積分は $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$ です:

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

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

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x} + \int{\tan{\left(x \right)} d x} = - \ln{\left(\left|{{\color{red}{\sin{\left(x \right)}}}}\right| \right)} - \int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x} + \int{\tan{\left(x \right)} d x}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\left(3 \int{\tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x}\right)}}$$

被積分関数を書き換える:

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

正接を用いて書き換えよ:

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

$$$u=3 x$$$ とする。

すると $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{3}$$$ となります。

積分は次のようになります

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

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{3}$$$ と $$$f{\left(u \right)} = \tan^{2}{\left(u \right)}$$$ に対して適用する:

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

$$$v=\tan{\left(u \right)}$$$ とする。

すると $$$u=\operatorname{atan}{\left(v \right)}$$$ および $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$（手順は»で確認できます）。

したがって、

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

分数を変形して分解する:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}}$$

項別に積分せよ:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\left(\int{1 d v} - \int{\frac{1}{v^{2} + 1} d v}\right)}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + \int{\frac{1}{v^{2} + 1} d v} - {\color{red}{\int{1 d v}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + \int{\frac{1}{v^{2} + 1} d v} - {\color{red}{v}}$$

$$$\frac{1}{v^{2} + 1}$$$ の不定積分は $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$ です:

$$- v - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}} = - v - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + {\color{red}{\operatorname{atan}{\left(v \right)}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + \operatorname{atan}{\left({\color{red}{v}} \right)} - {\color{red}{v}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + \operatorname{atan}{\left({\color{red}{\tan{\left(u \right)}}} \right)} - {\color{red}{\tan{\left(u \right)}}}$$

次のことを思い出してください $$$u=3 x$$$:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - \tan{\left({\color{red}{u}} \right)} + \operatorname{atan}{\left(\tan{\left({\color{red}{u}} \right)} \right)} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - \tan{\left({\color{red}{\left(3 x\right)}} \right)} + \operatorname{atan}{\left(\tan{\left({\color{red}{\left(3 x\right)}} \right)} \right)}$$

正接を$$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$に書き換える:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\tan{\left(x \right)} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

$$$u=\cos{\left(x \right)}$$$ 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、

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

$$$\frac{1}{u}$$$ の不定積分は $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$ です:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)}$$

したがって、

$$\int{\left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)d x} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)}$$

簡単化せよ:

$$\int{\left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)d x} = 3 x - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)}$$

積分定数を加える:

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

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