Integral dari $$$\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx$$$.
Solusi
Misalkan $$$u=\sin{\left(2 x \right)}$$$.
Kemudian $$$du=\left(\sin{\left(2 x \right)}\right)^{\prime }dx = 2 \cos{\left(2 x \right)} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$.
Jadi,
$${\color{red}{\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 u} d u}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \frac{1}{u}$$$:
$${\color{red}{\int{\frac{1}{2 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}$$
Integral dari $$$\frac{1}{u}$$$ adalah $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
Ingat bahwa $$$u=\sin{\left(2 x \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\sin{\left(2 x \right)}}}}\right| \right)}}{2}$$
Oleh karena itu,
$$\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{2}$$
Tambahkan konstanta integrasi:
$$\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{2}+C$$
Jawaban
$$$\int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx = \frac{\ln\left(\left|{\sin{\left(2 x \right)}}\right|\right)}{2} + C$$$A