$$$\sec^{3}{\left(\theta \right)}$$$'nin integrali

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

$$$\int{\sec^{3}{\left(\theta \right)} d \theta}$$$ integrali için, kısmi integrasyonu $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$ kullanın.

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

O halde $$$\operatorname{du}=\left(\sec{\left(\theta \right)}\right)^{\prime }d\theta=\tan{\left(\theta \right)} \sec{\left(\theta \right)} d\theta$$$ (adımlar için bkz. ») ve $$$\operatorname{v}=\int{\sec^{2}{\left(\theta \right)} d \theta}=\tan{\left(\theta \right)}$$$ (adımlar için bkz. »).

O halde,

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

Formülü $$$\tan^{2}{\left(\theta \right)} = \sec^{2}{\left(\theta \right)} - 1$$$ uygulayın:

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

Genişletin:

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

Farkın integrali, integrallerin farkıdır:

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

Dolayısıyla, integrale göre aşağıdaki basit doğrusal denklemi elde ederiz:

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

Bunu çözdüğümüzde, şunu elde ederiz:

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

Sekantı $$$\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$$$ olarak yeniden yazın:

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

Kosinüsü, $$$\cos\left(\theta\right)=\sin\left(\theta + \frac{\pi}{2}\right)$$$ formülünü kullanarak sinüs cinsinden yeniden yazın ve ardından sinüsü çift açı formülü $$$\sin\left(\theta\right)=2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$$$ kullanarak yeniden yazın.:

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

Payı ve paydayı $$$\sec^2\left(\frac{\theta}{2} + \frac{\pi}{4} \right)$$$ ile çarpın.:

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

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

Böylece $$$du=\left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)^{\prime }d\theta = \frac{\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}{2} d\theta$$$ (adımlar » görülebilir) ve $$$\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} d\theta = 2 du$$$ elde ederiz.

O halde,

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

$$$\frac{1}{u}$$$'nin integrali $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Hatırlayın ki $$$u=\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$:

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

Dolayısıyla,

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

İntegrasyon sabitini ekleyin:

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

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