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

$$$\int \sec^{3}{\left(\theta \right)}\, d\theta$$$ を求めよ。

積分 $$$\int{\sec^{3}{\left(\theta \right)} d \theta}$$$ には、部分積分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$を用いてください。

$$$\operatorname{u}=\sec{\left(\theta \right)}$$$ と $$$\operatorname{dv}=\sec^{2}{\left(\theta \right)} d\theta$$$ とする。

したがって、$$$\operatorname{du}=\left(\sec{\left(\theta \right)}\right)^{\prime }d\theta=\tan{\left(\theta \right)} \sec{\left(\theta \right)} d\theta$$$（手順は»を参照）および$$$\operatorname{v}=\int{\sec^{2}{\left(\theta \right)} d \theta}=\tan{\left(\theta \right)}$$$（手順は»を参照）。

積分は次のようになります

$$\int{\sec^{3}{\left(\theta \right)} d \theta}=\sec{\left(\theta \right)} \cdot \tan{\left(\theta \right)}-\int{\tan{\left(\theta \right)} \cdot \tan{\left(\theta \right)} \sec{\left(\theta \right)} d \theta}=\tan{\left(\theta \right)} \sec{\left(\theta \right)} - \int{\tan^{2}{\left(\theta \right)} \sec{\left(\theta \right)} d \theta}$$

公式$$$\tan^{2}{\left(\theta \right)} = \sec^{2}{\left(\theta \right)} - 1$$$を適用します:

$$\tan{\left(\theta \right)} \sec{\left(\theta \right)} - \int{\tan^{2}{\left(\theta \right)} \sec{\left(\theta \right)} d \theta}=\tan{\left(\theta \right)} \sec{\left(\theta \right)} - \int{\left(\sec^{2}{\left(\theta \right)} - 1\right) \sec{\left(\theta \right)} d \theta}$$

展開せよ:

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

差の積分は積分の差に等しい:

$$\tan{\left(\theta \right)} \sec{\left(\theta \right)} - \int{\left(\sec^{3}{\left(\theta \right)} - \sec{\left(\theta \right)}\right)d \theta}=\tan{\left(\theta \right)} \sec{\left(\theta \right)} + \int{\sec{\left(\theta \right)} d \theta} - \int{\sec^{3}{\left(\theta \right)} d \theta}$$

したがって、積分に関する次の簡単な線形方程式が得られます。

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

これを解くと、次のようになる。

$$\int{\sec^{3}{\left(\theta \right)} d \theta}=\frac{\tan{\left(\theta \right)} \sec{\left(\theta \right)}}{2} + \frac{\int{\sec{\left(\theta \right)} d \theta}}{2}$$

正割関数を$$$\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$$$として書き換える:

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

公式 $$$\cos\left(\theta\right)=\sin\left(\theta + \frac{\pi}{2}\right)$$$ を用いて余弦を正弦で表し、次に2倍角の公式 $$$\sin\left(\theta\right)=2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$$$ を用いて正弦を書き換えなさい。:

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

分子と分母に$$$\sec^2\left(\frac{\theta}{2} + \frac{\pi}{4} \right)$$$を掛ける:

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

$$$u=\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$ とする。

すると $$$du=\left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)^{\prime }d\theta = \frac{\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}{2} d\theta$$$（手順は»で確認できます）、$$$\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} d\theta = 2 du$$$ となります。

したがって、

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

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

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

次のことを思い出してください $$$u=\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + \frac{\tan{\left(\theta \right)} \sec{\left(\theta \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{2} + \frac{\tan{\left(\theta \right)} \sec{\left(\theta \right)}}{2}$$

したがって、

$$\int{\sec^{3}{\left(\theta \right)} d \theta} = \frac{\ln{\left(\left|{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{2} + \frac{\tan{\left(\theta \right)} \sec{\left(\theta \right)}}{2}$$

積分定数を加える:

$$\int{\sec^{3}{\left(\theta \right)} d \theta} = \frac{\ln{\left(\left|{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{2} + \frac{\tan{\left(\theta \right)} \sec{\left(\theta \right)}}{2}+C$$

$$$\int \sec^{3}{\left(\theta \right)}\, d\theta = \left(\frac{\ln\left(\left|{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}\right|\right)}{2} + \frac{\tan{\left(\theta \right)} \sec{\left(\theta \right)}}{2}\right) + C$$$A