Integral dari $$$\sin^{4}{\left(6 x \right)}$$$

Temukan $$$\int \sin^{4}{\left(6 x \right)}\, dx$$$.

Misalkan $$$u=6 x$$$.

Kemudian $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{6}$$$.

Oleh karena itu,

$${\color{red}{\int{\sin^{4}{\left(6 x \right)} d x}}} = {\color{red}{\int{\frac{\sin^{4}{\left(u \right)}}{6} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{6}$$$ dan $$$f{\left(u \right)} = \sin^{4}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sin^{4}{\left(u \right)}}{6} d u}}} = {\color{red}{\left(\frac{\int{\sin^{4}{\left(u \right)} d u}}{6}\right)}}$$

Terapkan rumus reduksi pangkat $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ dengan $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\sin^{4}{\left(u \right)} d u}}}}{6} = \frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{6}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{8}$$$ dan $$$f{\left(u \right)} = - 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$:

$$\frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{6} = \frac{{\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}{8}\right)}}}{6}$$

Integralkan suku demi suku:

$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}}}{48} = \frac{{\color{red}{\left(\int{3 d u} - \int{4 \cos{\left(2 u \right)} d u} + \int{\cos{\left(4 u \right)} d u}\right)}}}{48}$$

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=3$$$:

$$- \frac{\int{4 \cos{\left(2 u \right)} d u}}{48} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} + \frac{{\color{red}{\int{3 d u}}}}{48} = - \frac{\int{4 \cos{\left(2 u \right)} d u}}{48} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} + \frac{{\color{red}{\left(3 u\right)}}}{48}$$

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

$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{4 \cos{\left(2 u \right)} d u}}}}{48} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 u \right)} d u}\right)}}}{48}$$

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

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

Integralnya menjadi

$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{12} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{12}$$

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

$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{12} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{12}$$

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

$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{24} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{{\color{red}{\sin{\left(v \right)}}}}{24}$$

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

$$\frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{\sin{\left({\color{red}{v}} \right)}}{24} = \frac{u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{48} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{24}$$

Misalkan $$$v=4 u$$$.

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

Dengan demikian,

$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{48} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{48}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{48} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{48}$$

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

$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{192} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{{\color{red}{\sin{\left(v \right)}}}}{192}$$

Ingat bahwa $$$v=4 u$$$:

$$\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{\sin{\left({\color{red}{v}} \right)}}{192} = \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{24} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{192}$$

Ingat bahwa $$$u=6 x$$$:

$$- \frac{\sin{\left(2 {\color{red}{u}} \right)}}{24} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{192} + \frac{{\color{red}{u}}}{16} = - \frac{\sin{\left(2 {\color{red}{\left(6 x\right)}} \right)}}{24} + \frac{\sin{\left(4 {\color{red}{\left(6 x\right)}} \right)}}{192} + \frac{{\color{red}{\left(6 x\right)}}}{16}$$

Oleh karena itu,

$$\int{\sin^{4}{\left(6 x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(12 x \right)}}{24} + \frac{\sin{\left(24 x \right)}}{192}$$

Sederhanakan:

$$\int{\sin^{4}{\left(6 x \right)} d x} = \frac{72 x - 8 \sin{\left(12 x \right)} + \sin{\left(24 x \right)}}{192}$$

Tambahkan konstanta integrasi:

$$\int{\sin^{4}{\left(6 x \right)} d x} = \frac{72 x - 8 \sin{\left(12 x \right)} + \sin{\left(24 x \right)}}{192}+C$$

$$$\int \sin^{4}{\left(6 x \right)}\, dx = \frac{72 x - 8 \sin{\left(12 x \right)} + \sin{\left(24 x \right)}}{192} + C$$$A