$$$\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20}$$$ 的積分
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您的輸入
求$$$\int \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20}\, dt$$$。
解答
使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$,以 $$$\alpha=2 t$$$ 和 $$$\beta=4 t$$$ 將 $$$\sin\left(2 t \right)\cos\left(4 t \right)$$$ 改寫:
$${\color{red}{\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t}}} = {\color{red}{\int{\frac{\pi \left(- \frac{\sin{\left(2 t \right)}}{2} + \frac{\sin{\left(6 t \right)}}{2}\right) \sin{\left(4 t \right)}}{20} d t}}}$$
展開該表達式:
$${\color{red}{\int{\frac{\pi \left(- \frac{\sin{\left(2 t \right)}}{2} + \frac{\sin{\left(6 t \right)}}{2}\right) \sin{\left(4 t \right)}}{20} d t}}} = {\color{red}{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{40} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{40}\right)d t}}}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(t \right)} = - \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20}$$$:
$${\color{red}{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{40} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{40}\right)d t}}} = {\color{red}{\left(\frac{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20}\right)d t}}{2}\right)}}$$
逐項積分:
$$\frac{{\color{red}{\int{\left(- \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} + \frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20}\right)d t}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} d t} + \int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}\right)}}}{2}$$
使用公式 $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$,以 $$$\alpha=2 t$$$ 和 $$$\beta=4 t$$$ 將 $$$\sin\left(2 t \right)\sin\left(4 t \right)$$$ 改寫:
$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)}}{20} d t}}}}{2} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)}{20} d t}}}}{2}$$
展開該表達式:
$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)}{20} d t}}}}{2} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(6 t \right)}}{40}\right)d t}}}}{2}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(t \right)} = \frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(6 t \right)}}{20}$$$:
$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(6 t \right)}}{40}\right)d t}}}}{2} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\left(\frac{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(6 t \right)}}{20}\right)d t}}{2}\right)}}}{2}$$
逐項積分:
$$\frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(6 t \right)}}{20}\right)d t}}}}{4} = \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\left(\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t} - \int{\frac{\pi \cos{\left(6 t \right)}}{20} d t}\right)}}}{4}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{\pi}{20}$$$ 與 $$$f{\left(t \right)} = \cos{\left(6 t \right)}$$$:
$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{{\color{red}{\int{\frac{\pi \cos{\left(6 t \right)}}{20} d t}}}}{4} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(6 t \right)} d t}}{20}\right)}}}{4}$$
令 $$$u=6 t$$$。
則 $$$du=\left(6 t\right)^{\prime }dt = 6 dt$$$ (步驟見»),並可得 $$$dt = \frac{du}{6}$$$。
因此,
$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\cos{\left(6 t \right)} d t}}}}{80} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{80}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{6}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{80} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{80}$$
餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\int{\cos{\left(u \right)} d u}}}}{480} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi {\color{red}{\sin{\left(u \right)}}}}{480}$$
回顧一下 $$$u=6 t$$$:
$$- \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi \sin{\left({\color{red}{u}} \right)}}{480} = - \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} + \frac{\pi \sin{\left({\color{red}{\left(6 t\right)}} \right)}}{480}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{\pi}{20}$$$ 與 $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$:
$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}}}{4} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(2 t \right)} d t}}{20}\right)}}}{4}$$
令 $$$u=2 t$$$。
則 $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (步驟見»),並可得 $$$dt = \frac{du}{2}$$$。
該積分變為
$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{80} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{80}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{80} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{80}$$
餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\int{\cos{\left(u \right)} d u}}}}{160} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi {\color{red}{\sin{\left(u \right)}}}}{160}$$
回顧一下 $$$u=2 t$$$:
$$\frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi \sin{\left({\color{red}{u}} \right)}}{160} = \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}{2} - \frac{\pi \sin{\left({\color{red}{\left(2 t\right)}} \right)}}{160}$$
使用公式 $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$,以 $$$\alpha=4 t$$$ 和 $$$\beta=6 t$$$ 將 $$$\sin\left(4 t \right)\sin\left(6 t \right)$$$ 改寫:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\frac{\pi \sin{\left(4 t \right)} \sin{\left(6 t \right)}}{20} d t}}}}{2} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(10 t \right)}}{2}\right)}{20} d t}}}}{2}$$
展開該表達式:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\frac{\pi \left(\frac{\cos{\left(2 t \right)}}{2} - \frac{\cos{\left(10 t \right)}}{2}\right)}{20} d t}}}}{2} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(10 t \right)}}{40}\right)d t}}}}{2}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(t \right)} = \frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(10 t \right)}}{20}$$$:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{40} - \frac{\pi \cos{\left(10 t \right)}}{40}\right)d t}}}}{2} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\left(\frac{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(10 t \right)}}{20}\right)d t}}{2}\right)}}}{2}$$
逐項積分:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\int{\left(\frac{\pi \cos{\left(2 t \right)}}{20} - \frac{\pi \cos{\left(10 t \right)}}{20}\right)d t}}}}{4} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{{\color{red}{\left(\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t} - \int{\frac{\pi \cos{\left(10 t \right)}}{20} d t}\right)}}}{4}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{\pi}{20}$$$ 與 $$$f{\left(t \right)} = \cos{\left(10 t \right)}$$$:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{{\color{red}{\int{\frac{\pi \cos{\left(10 t \right)}}{20} d t}}}}{4} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(10 t \right)} d t}}{20}\right)}}}{4}$$
令 $$$u=10 t$$$。
則 $$$du=\left(10 t\right)^{\prime }dt = 10 dt$$$ (步驟見»),並可得 $$$dt = \frac{du}{10}$$$。
所以,
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\cos{\left(10 t \right)} d t}}}}{80} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{10} d u}}}}{80}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{10}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\frac{\cos{\left(u \right)}}{10} d u}}}}{80} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{10}\right)}}}{80}$$
餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\int{\cos{\left(u \right)} d u}}}}{800} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi {\color{red}{\sin{\left(u \right)}}}}{800}$$
回顧一下 $$$u=10 t$$$:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi \sin{\left({\color{red}{u}} \right)}}{800} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} + \frac{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}{4} - \frac{\pi \sin{\left({\color{red}{\left(10 t\right)}} \right)}}{800}$$
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$,使用 $$$c=\frac{\pi}{20}$$$ 與 $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$:
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{{\color{red}{\int{\frac{\pi \cos{\left(2 t \right)}}{20} d t}}}}{4} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{{\color{red}{\left(\frac{\pi \int{\cos{\left(2 t \right)} d t}}{20}\right)}}}{4}$$
積分 $$$\int{\cos{\left(2 t \right)} d t}$$$ 已經計算過:
$$\int{\cos{\left(2 t \right)} d t} = \frac{\sin{\left(2 t \right)}}{2}$$
因此,
$$- \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{\pi {\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{80} = - \frac{\pi \sin{\left(2 t \right)}}{160} + \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800} + \frac{\pi {\color{red}{\left(\frac{\sin{\left(2 t \right)}}{2}\right)}}}{80}$$
因此,
$$\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t} = \frac{\pi \sin{\left(6 t \right)}}{480} - \frac{\pi \sin{\left(10 t \right)}}{800}$$
化簡:
$$\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t} = \frac{\pi \left(5 \sin{\left(6 t \right)} - 3 \sin{\left(10 t \right)}\right)}{2400}$$
加上積分常數:
$$\int{\frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20} d t} = \frac{\pi \left(5 \sin{\left(6 t \right)} - 3 \sin{\left(10 t \right)}\right)}{2400}+C$$
答案
$$$\int \frac{\pi \sin{\left(2 t \right)} \sin{\left(4 t \right)} \cos{\left(4 t \right)}}{20}\, dt = \frac{\pi \left(5 \sin{\left(6 t \right)} - 3 \sin{\left(10 t \right)}\right)}{2400} + C$$$A