$$$\tan{\left(x \right)} + 1 - \frac{1}{\tan{\left(x \right)}}$$$の積分

$$$\int \left(\tan{\left(x \right)} + 1 - \frac{1}{\tan{\left(x \right)}}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(\tan{\left(x \right)} + 1 - \frac{1}{\tan{\left(x \right)}}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{\frac{1}{\tan{\left(x \right)}} d x} + \int{\tan{\left(x \right)} d x}\right)}}$$

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

$$- \int{\frac{1}{\tan{\left(x \right)}} d x} + \int{\tan{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = - \int{\frac{1}{\tan{\left(x \right)}} d x} + \int{\tan{\left(x \right)} d x} + {\color{red}{x}}$$

$$$u=\tan{\left(x \right)}$$$ とする。

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

したがって、

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

$$$v=u^{2} + 1$$$ とする。

すると $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$（手順は»で確認できます）、$$$u du = \frac{dv}{2}$$$ となります。

したがって、

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

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

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

部分分数分解を行う (手順は»で確認できます):

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

項別に積分せよ:

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

$$$w=v - 1$$$ とする。

すると $$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$（手順は»で確認できます）、$$$dv = dw$$$ となります。

積分は次のようになります

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

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

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

次のことを思い出してください $$$w=v - 1$$$:

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

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

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

次のことを思い出してください $$$v=u^{2} + 1$$$:

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

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

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

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

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

$$$u=\cos{\left(x \right)}$$$ 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、

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

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:

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

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

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

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

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

$$\int{\left(\tan{\left(x \right)} + 1 - \frac{1}{\tan{\left(x \right)}}\right)d x} = x + \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} - \ln{\left(\tan{\left(x \right)} \right)} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$$

$$$\int \left(\tan{\left(x \right)} + 1 - \frac{1}{\tan{\left(x \right)}}\right)\, dx = \left(x + \frac{\ln\left(\tan^{2}{\left(x \right)} + 1\right)}{2} - \ln\left(\tan{\left(x \right)}\right) - \ln\left(\left|{\cos{\left(x \right)}}\right|\right)\right) + C$$$A