$$$\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$の積分

$$$\int \frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx$$$ を求めよ。

分子と分母に$$$\frac{1}{\cos^{2}{\left(x \right)}}$$$を掛け、$$$\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$を$$$\frac{1}{\tan^{2}{\left(x \right)}}$$$に変換します。:

$${\color{red}{\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}}$$

分子と分母に$$$\cos^{2}{\left(x \right)}$$$を掛け、$$$\frac{1}{\cos^{2}{\left(x \right)}}$$$を$$$\sec^{2}{\left(x \right)}$$$に変換します。:

$${\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos^{4}{\left(x \right)} \sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}}$$

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。:

$${\color{red}{\int{\frac{\cos^{4}{\left(x \right)} \sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\left(\tan^{2}{\left(x \right)} + 1\right)^{2} \tan^{2}{\left(x \right)}} d x}}}$$

$$$u=\tan{\left(x \right)}$$$ 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、

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

部分分数分解を行う (手順は»で確認できます):

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

項別に積分せよ:

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

$$$n=-2$$$ 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:

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

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

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

積分$$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}$$$を計算するには、積分$$$\int{\frac{1}{u^{2} + 1} d u}$$$に部分積分法$$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$を適用します。

$$$\operatorname{c}=\frac{1}{u^{2} + 1}$$$ と $$$\operatorname{dv}=du$$$ とする。

したがって、$$$\operatorname{dc}=\left(\frac{1}{u^{2} + 1}\right)^{\prime }du=- \frac{2 u}{\left(u^{2} + 1\right)^{2}} du$$$（手順は»を参照）および$$$\operatorname{v}=\int{1 d u}=u$$$（手順は»を参照）。

積分は次のようになります

$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}=\frac{1}{u^{2} + 1} \cdot u-\int{u \cdot \left(- \frac{2 u}{\left(u^{2} + 1\right)^{2}}\right) d u}=\frac{u}{u^{2} + 1} - \int{\left(- \frac{2 u^{2}}{\left(u^{2} + 1\right)^{2}}\right)d u}$$

定数をくくり出す:

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

被積分関数の分子を$$$u^{2}=u^{2}{\color{red}{+1}}{\color{red}{-1}}$$$と書き換えて、分割する：

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

積分を分割する:

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

したがって、積分に関する次の簡単な一次方程式を得ます:

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

これを解くと、次のようになる

$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}=\frac{u}{2 \left(u^{2} + 1\right)} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{2}$$

したがって、

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

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

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

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

$$- \frac{3 \operatorname{atan}{\left({\color{red}{u}} \right)}}{2} - {\color{red}{u}}^{-1} - \frac{{\color{red}{u}} \left(1 + {\color{red}{u}}^{2}\right)^{-1}}{2} = - \frac{3 \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)}}{2} - {\color{red}{\tan{\left(x \right)}}}^{-1} - \frac{{\color{red}{\tan{\left(x \right)}}} \left(1 + {\color{red}{\tan{\left(x \right)}}}^{2}\right)^{-1}}{2}$$

したがって、

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 \operatorname{atan}{\left(\tan{\left(x \right)} \right)}}{2} - \frac{1}{\tan{\left(x \right)}} - \frac{\tan{\left(x \right)}}{2 \left(\tan^{2}{\left(x \right)} + 1\right)}$$

簡単化せよ:

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}$$

積分定数を加える:

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

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