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