$$$\frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2}$$$ 的積分
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您的輸入
求$$$\int \frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2}\, dx$$$。
解答
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \sin{\left(2 x \right)} \cos{\left(2 x \right)}$$$:
$${\color{red}{\int{\frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x}}{2}\right)}}$$
令 $$$u=\sin{\left(2 x \right)}$$$。
則 $$$du=\left(\sin{\left(2 x \right)}\right)^{\prime }dx = 2 \cos{\left(2 x \right)} dx$$$ (步驟見»),並可得 $$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$。
因此,
$$\frac{{\color{red}{\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x}}}}{2} = \frac{{\color{red}{\int{\frac{u}{2} d u}}}}{2}$$
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 $$$f{\left(u \right)} = u$$$:
$$\frac{{\color{red}{\int{\frac{u}{2} d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{u d u}}{2}\right)}}}{2}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=1$$$:
$$\frac{{\color{red}{\int{u d u}}}}{4}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{4}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{4}$$
回顧一下 $$$u=\sin{\left(2 x \right)}$$$:
$$\frac{{\color{red}{u}}^{2}}{8} = \frac{{\color{red}{\sin{\left(2 x \right)}}}^{2}}{8}$$
因此,
$$\int{\frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2} d x} = \frac{\sin^{2}{\left(2 x \right)}}{8}$$
加上積分常數:
$$\int{\frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2} d x} = \frac{\sin^{2}{\left(2 x \right)}}{8}+C$$
答案
$$$\int \frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2}\, dx = \frac{\sin^{2}{\left(2 x \right)}}{8} + C$$$A