Integral dari $$$- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}$$$

Temukan $$$\int \left(- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}\right)\, dx$$$.

Integralkan suku demi suku:

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

Tulis ulang kosinus dalam bentuk sinus menggunakan rumus $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ dan kemudian tulis ulang sinus menggunakan rumus sudut rangkap $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

Kalikan pembilang dan penyebut dengan $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

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

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

Kemudian $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

Dengan demikian,

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

Integral dari $$$\frac{1}{u}$$$ adalah $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

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

Integral dari sinus adalah $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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