Integral dari $$$\frac{\sin{\left(x \right)}}{\cos^{3}{\left(x \right)}}$$$

Temukan $$$\int \frac{\sin{\left(x \right)}}{\cos^{3}{\left(x \right)}}\, dx$$$.

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

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

Integral tersebut dapat ditulis ulang sebagai

$${\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos^{3}{\left(x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{1}{u^{3}}\right)d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = \frac{1}{u^{3}}$$$:

$${\color{red}{\int{\left(- \frac{1}{u^{3}}\right)d u}}} = {\color{red}{\left(- \int{\frac{1}{u^{3}} d u}\right)}}$$

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-3$$$:

$$- {\color{red}{\int{\frac{1}{u^{3}} d u}}}=- {\color{red}{\int{u^{-3} d u}}}=- {\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}=- {\color{red}{\left(- \frac{u^{-2}}{2}\right)}}=- {\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$

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

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

Oleh karena itu,

$$\int{\frac{\sin{\left(x \right)}}{\cos^{3}{\left(x \right)}} d x} = \frac{1}{2 \cos^{2}{\left(x \right)}}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sin{\left(x \right)}}{\cos^{3}{\left(x \right)}} d x} = \frac{1}{2 \cos^{2}{\left(x \right)}}+C$$

$$$\int \frac{\sin{\left(x \right)}}{\cos^{3}{\left(x \right)}}\, dx = \frac{1}{2 \cos^{2}{\left(x \right)}} + C$$$A