Integral dari $$$\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}$$$

Temukan $$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx$$$.

Misalkan $$$u=8 x$$$.

Kemudian $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{8}$$$.

Integralnya menjadi

$${\color{red}{\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{8}$$$ dan $$$f{\left(u \right)} = \frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}}$$$:

$${\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}} d u}}{8}\right)}}$$

Tulis ulang integran:

$$\frac{{\color{red}{\int{\frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}} d u}}}}{8} = \frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=2$$$ dan $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8} = \frac{{\color{red}{\left(2 \int{\sin{\left(u \right)} d u}\right)}}}{8}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$

Ingat bahwa $$$u=8 x$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{4} = - \frac{\cos{\left({\color{red}{\left(8 x\right)}} \right)}}{4}$$

Oleh karena itu,

$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}+C$$

$$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx = - \frac{\cos{\left(8 x \right)}}{4} + C$$$A