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

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

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

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

Integralnya menjadi

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

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

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

Terapkan rumus reduksi pangkat $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ dengan $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\sin^{2}{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{4}$$

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

$$\frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}{2}\right)}}}{4}$$

Integralkan suku demi suku:

$$\frac{{\color{red}{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}}}{8} = \frac{{\color{red}{\left(\int{1 d u} - \int{\cos{\left(2 u \right)} d u}\right)}}}{8}$$

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:

$$- \frac{\int{\cos{\left(2 u \right)} d u}}{8} + \frac{{\color{red}{\int{1 d u}}}}{8} = - \frac{\int{\cos{\left(2 u \right)} d u}}{8} + \frac{{\color{red}{u}}}{8}$$

Misalkan $$$v=2 u$$$.

Kemudian $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = \frac{dv}{2}$$$.

Jadi,

$$\frac{u}{8} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{8} = \frac{u}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{8}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{u}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{8} = \frac{u}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{8}$$

Integral dari kosinus adalah $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{u}{8} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{16} = \frac{u}{8} - \frac{{\color{red}{\sin{\left(v \right)}}}}{16}$$

Ingat bahwa $$$v=2 u$$$:

$$\frac{u}{8} - \frac{\sin{\left({\color{red}{v}} \right)}}{16} = \frac{u}{8} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{16}$$

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

$$- \frac{\sin{\left(2 {\color{red}{u}} \right)}}{16} + \frac{{\color{red}{u}}}{8} = - \frac{\sin{\left(2 {\color{red}{\left(4 x\right)}} \right)}}{16} + \frac{{\color{red}{\left(4 x\right)}}}{8}$$

Oleh karena itu,

$$\int{\sin^{2}{\left(4 x \right)} d x} = \frac{x}{2} - \frac{\sin{\left(8 x \right)}}{16}$$

Tambahkan konstanta integrasi:

$$\int{\sin^{2}{\left(4 x \right)} d x} = \frac{x}{2} - \frac{\sin{\left(8 x \right)}}{16}+C$$

$$$\int \sin^{2}{\left(4 x \right)}\, dx = \left(\frac{x}{2} - \frac{\sin{\left(8 x \right)}}{16}\right) + C$$$A