Integraal van $$$\sec^{3}{\left(\theta \right)}$$$

Bepaal $$$\int \sec^{3}{\left(\theta \right)}\, d\theta$$$.

Voor de integraal $$$\int{\sec^{3}{\left(\theta \right)} d \theta}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Zij $$$\operatorname{u}=\sec{\left(\theta \right)}$$$ en $$$\operatorname{dv}=\sec^{2}{\left(\theta \right)} d\theta$$$.

Dan $$$\operatorname{du}=\left(\sec{\left(\theta \right)}\right)^{\prime }d\theta=\tan{\left(\theta \right)} \sec{\left(\theta \right)} d\theta$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{\sec^{2}{\left(\theta \right)} d \theta}=\tan{\left(\theta \right)}$$$ (de stappen zijn te zien »).

Dus,

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

Pas de formule $$$\tan^{2}{\left(\theta \right)} = \sec^{2}{\left(\theta \right)} - 1$$$ toe:

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

Werk uit:

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

De integraal van een verschil is het verschil van integralen:

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

Dus krijgen we de volgende eenvoudige lineaire vergelijking met betrekking tot de integraal:

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

Door het op te lossen, verkrijgen we dat

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

Herschrijf de secans als $$$\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$$$:

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

Herschrijf de cosinus in termen van de sinus met behulp van de formule $$$\cos\left(\theta\right)=\sin\left(\theta + \frac{\pi}{2}\right)$$$ en herschrijf vervolgens de sinus met behulp van de dubbelhoekformule $$$\sin\left(\theta\right)=2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$$$:

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

Vermenigvuldig de teller en de noemer met $$$\sec^2\left(\frac{\theta}{2} + \frac{\pi}{4} \right)$$$:

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

Zij $$$u=\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$.

Dan $$$du=\left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)^{\prime }d\theta = \frac{\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}{2} d\theta$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} d\theta = 2 du$$$.

Dus,

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

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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