Integral dari $$$\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}\, d\theta$$$.
Solusi
Tulis ulang integran menggunakan rumus sudut rangkap $$$\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1}{2}\sin\left( 2 \theta \right)$$$:
$${\color{red}{\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\sin^{6}{\left(2 \theta \right)}}{64} d \theta}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ dengan $$$c=\frac{1}{64}$$$ dan $$$f{\left(\theta \right)} = \sin^{6}{\left(2 \theta \right)}$$$:
$${\color{red}{\int{\frac{\sin^{6}{\left(2 \theta \right)}}{64} d \theta}}} = {\color{red}{\left(\frac{\int{\sin^{6}{\left(2 \theta \right)} d \theta}}{64}\right)}}$$
Terapkan rumus reduksi pangkat $$$\sin^{6}{\left(\alpha \right)} = - \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} - \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$$$ dengan $$$\alpha=2 \theta$$$:
$$\frac{{\color{red}{\int{\sin^{6}{\left(2 \theta \right)} d \theta}}}}{64} = \frac{{\color{red}{\int{\left(- \frac{15 \cos{\left(4 \theta \right)}}{32} + \frac{3 \cos{\left(8 \theta \right)}}{16} - \frac{\cos{\left(12 \theta \right)}}{32} + \frac{5}{16}\right)d \theta}}}}{64}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ dengan $$$c=\frac{1}{32}$$$ dan $$$f{\left(\theta \right)} = - 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10$$$:
$$\frac{{\color{red}{\int{\left(- \frac{15 \cos{\left(4 \theta \right)}}{32} + \frac{3 \cos{\left(8 \theta \right)}}{16} - \frac{\cos{\left(12 \theta \right)}}{32} + \frac{5}{16}\right)d \theta}}}}{64} = \frac{{\color{red}{\left(\frac{\int{\left(- 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10\right)d \theta}}{32}\right)}}}{64}$$
Integralkan suku demi suku:
$$\frac{{\color{red}{\int{\left(- 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10\right)d \theta}}}}{2048} = \frac{{\color{red}{\left(\int{10 d \theta} - \int{15 \cos{\left(4 \theta \right)} d \theta} + \int{6 \cos{\left(8 \theta \right)} d \theta} - \int{\cos{\left(12 \theta \right)} d \theta}\right)}}}{2048}$$
Terapkan aturan konstanta $$$\int c\, d\theta = c \theta$$$ dengan $$$c=10$$$:
$$- \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\int{\cos{\left(12 \theta \right)} d \theta}}{2048} + \frac{{\color{red}{\int{10 d \theta}}}}{2048} = - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\int{\cos{\left(12 \theta \right)} d \theta}}{2048} + \frac{{\color{red}{\left(10 \theta\right)}}}{2048}$$
Misalkan $$$u=12 \theta$$$.
Kemudian $$$du=\left(12 \theta\right)^{\prime }d\theta = 12 d\theta$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$d\theta = \frac{du}{12}$$$.
Oleh karena itu,
$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\cos{\left(12 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{12} d u}}}}{2048}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{12}$$$ dan $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{12} d u}}}}{2048} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{12}\right)}}}{2048}$$
Integral dari kosinus adalah $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{24576} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\sin{\left(u \right)}}}}{24576}$$
Ingat bahwa $$$u=12 \theta$$$:
$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\sin{\left({\color{red}{u}} \right)}}{24576} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\sin{\left({\color{red}{\left(12 \theta\right)}} \right)}}{24576}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ dengan $$$c=15$$$ dan $$$f{\left(\theta \right)} = \cos{\left(4 \theta \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{15 \cos{\left(4 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\left(15 \int{\cos{\left(4 \theta \right)} d \theta}\right)}}}{2048}$$
Misalkan $$$u=4 \theta$$$.
Kemudian $$$du=\left(4 \theta\right)^{\prime }d\theta = 4 d\theta$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$d\theta = \frac{du}{4}$$$.
Dengan demikian,
$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2048}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2048}$$
Integral dari kosinus adalah $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8192} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\sin{\left(u \right)}}}}{8192}$$
Ingat bahwa $$$u=4 \theta$$$:
$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 \sin{\left({\color{red}{u}} \right)}}{8192} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 \sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{8192}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ dengan $$$c=6$$$ dan $$$f{\left(\theta \right)} = \cos{\left(8 \theta \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{{\color{red}{\int{6 \cos{\left(8 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{{\color{red}{\left(6 \int{\cos{\left(8 \theta \right)} d \theta}\right)}}}{2048}$$
Misalkan $$$u=8 \theta$$$.
Kemudian $$$du=\left(8 \theta\right)^{\prime }d\theta = 8 d\theta$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$d\theta = \frac{du}{8}$$$.
Integral tersebut dapat ditulis ulang sebagai
$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\cos{\left(8 \theta \right)} d \theta}}}}{1024} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{1024}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{8}$$$ dan $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{1024} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{1024}$$
Integral dari kosinus adalah $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8192} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{8192}$$
Ingat bahwa $$$u=8 \theta$$$:
$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{8192} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 \sin{\left({\color{red}{\left(8 \theta\right)}} \right)}}{8192}$$
Oleh karena itu,
$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} + \frac{3 \sin{\left(8 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576}$$
Sederhanakan:
$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576}$$
Tambahkan konstanta integrasi:
$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576}+C$$
Jawaban
$$$\int \sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}\, d\theta = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576} + C$$$A