Integral dari $$$- \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$

Temukan $$$\int \left(- \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}\right)\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=-1$$$ dan $$$f{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$:

$${\color{red}{\int{\left(- \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}\right)d x}}} = {\color{red}{\left(- \int{\frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}\right)}}$$

Misalkan $$$u=\sin{\left(x \right)}$$$.

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

Integral tersebut dapat ditulis ulang sebagai

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

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-2$$$:

$$- {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- {\color{red}{\int{u^{-2} d u}}}=- {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- {\color{red}{\left(- u^{-1}\right)}}=- {\color{red}{\left(- \frac{1}{u}\right)}}$$

Ingat bahwa $$$u=\sin{\left(x \right)}$$$:

$${\color{red}{u}}^{-1} = {\color{red}{\sin{\left(x \right)}}}^{-1}$$

Oleh karena itu,

$$\int{\left(- \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}\right)d x} = \frac{1}{\sin{\left(x \right)}}$$

Tambahkan konstanta integrasi:

$$\int{\left(- \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}\right)d x} = \frac{1}{\sin{\left(x \right)}}+C$$

$$$\int \left(- \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}\right)\, dx = \frac{1}{\sin{\left(x \right)}} + C$$$A