Integral dari $$$\tan^{2}{\left(4 x \right)}$$$

Temukan $$$\int \tan^{2}{\left(4 x \right)}\, dx$$$.

Misalkan $$$u=4 x$$$.

Kemudian $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{4}$$$.

Jadi,

$${\color{red}{\int{\tan^{2}{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\tan^{2}{\left(u \right)}}{4} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \tan^{2}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\tan^{2}{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\tan^{2}{\left(u \right)} d u}}{4}\right)}}$$

Misalkan $$$v=\tan{\left(u \right)}$$$.

Kemudian $$$u=\operatorname{atan}{\left(v \right)}$$$ dan $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$ (langkah-langkahnya dapat dilihat »).

Oleh karena itu,

$$\frac{{\color{red}{\int{\tan^{2}{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}}}{4}$$

Tulis ulang dan pisahkan pecahannya:

$$\frac{{\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}}}{4} = \frac{{\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}}}{4}$$

Integralkan suku demi suku:

$$\frac{{\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}}}{4} = \frac{{\color{red}{\left(\int{1 d v} - \int{\frac{1}{v^{2} + 1} d v}\right)}}}{4}$$

Terapkan aturan konstanta $$$\int c\, dv = c v$$$ dengan $$$c=1$$$:

$$- \frac{\int{\frac{1}{v^{2} + 1} d v}}{4} + \frac{{\color{red}{\int{1 d v}}}}{4} = - \frac{\int{\frac{1}{v^{2} + 1} d v}}{4} + \frac{{\color{red}{v}}}{4}$$

Integral dari $$$\frac{1}{v^{2} + 1}$$$ adalah $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$\frac{v}{4} - \frac{{\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{4} = \frac{v}{4} - \frac{{\color{red}{\operatorname{atan}{\left(v \right)}}}}{4}$$

Ingat bahwa $$$v=\tan{\left(u \right)}$$$:

$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{4} + \frac{{\color{red}{v}}}{4} = - \frac{\operatorname{atan}{\left({\color{red}{\tan{\left(u \right)}}} \right)}}{4} + \frac{{\color{red}{\tan{\left(u \right)}}}}{4}$$

Ingat bahwa $$$u=4 x$$$:

$$\frac{\tan{\left({\color{red}{u}} \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left({\color{red}{u}} \right)} \right)}}{4} = \frac{\tan{\left({\color{red}{\left(4 x\right)}} \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left({\color{red}{\left(4 x\right)}} \right)} \right)}}{4}$$

Oleh karena itu,

$$\int{\tan^{2}{\left(4 x \right)} d x} = \frac{\tan{\left(4 x \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left(4 x \right)} \right)}}{4}$$

Sederhanakan:

$$\int{\tan^{2}{\left(4 x \right)} d x} = - x + \frac{\tan{\left(4 x \right)}}{4}$$

Tambahkan konstanta integrasi:

$$\int{\tan^{2}{\left(4 x \right)} d x} = - x + \frac{\tan{\left(4 x \right)}}{4}+C$$

$$$\int \tan^{2}{\left(4 x \right)}\, dx = \left(- x + \frac{\tan{\left(4 x \right)}}{4}\right) + C$$$A