$$$- \tan{\left(x \right)} + \sec{\left(x \right)}$$$'nin integrali

Bulun: $$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)\, dx$$$.

Her terimin integralini alın:

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

Teğeti $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$ olarak yeniden yazın:

$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\tan{\left(x \right)} d x}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

$$$u=\cos{\left(x \right)}$$$ olsun.

Böylece $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (adımlar » görülebilir) ve $$$\sin{\left(x \right)} dx = - du$$$ elde ederiz.

Dolayısıyla,

$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=-1$$$ ve $$$f{\left(u \right)} = \frac{1}{u}$$$ ile uygula:

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

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

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

Hatırlayın ki $$$u=\cos{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\sec{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} + \int{\sec{\left(x \right)} d x}$$

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

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\sec{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}$$

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

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$

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

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

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

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

O halde,

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

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

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

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

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

Dolayısıyla,

$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}$$

İntegrasyon sabitini ekleyin:

$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$

$$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)\, dx = \left(\ln\left(\left|{\cos{\left(x \right)}}\right|\right) + \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)\right) + C$$$A