$$$4 \cos^{2}{\left(x \right)} - 1$$$ 的積分
您的輸入
求$$$\int \left(4 \cos^{2}{\left(x \right)} - 1\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(4 \cos^{2}{\left(x \right)} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{4 \cos^{2}{\left(x \right)} d x}\right)}}$$
配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$\int{4 \cos^{2}{\left(x \right)} d x} - {\color{red}{\int{1 d x}}} = \int{4 \cos^{2}{\left(x \right)} d x} - {\color{red}{x}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$：
$$- x + {\color{red}{\int{4 \cos^{2}{\left(x \right)} d x}}} = - x + {\color{red}{\left(4 \int{\cos^{2}{\left(x \right)} d x}\right)}}$$
套用降冪公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，令 $$$\alpha=x$$$:
$$- x + 4 {\color{red}{\int{\cos^{2}{\left(x \right)} d x}}} = - x + 4 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}$$
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 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$：
$$- x + 4 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}} = - x + 4 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}$$
逐項積分：
$$- x + 2 {\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}} = - x + 2 {\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}$$
配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$- x + 2 \int{\cos{\left(2 x \right)} d x} + 2 {\color{red}{\int{1 d x}}} = - x + 2 \int{\cos{\left(2 x \right)} d x} + 2 {\color{red}{x}}$$
令 $$$u=2 x$$$。
則 $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。
該積分變為
$$x + 2 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = x + 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$
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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：
$$x + 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = x + 2 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$
餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：
$$x + {\color{red}{\int{\cos{\left(u \right)} d u}}} = x + {\color{red}{\sin{\left(u \right)}}}$$
回顧一下 $$$u=2 x$$$：
$$x + \sin{\left({\color{red}{u}} \right)} = x + \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$
因此，
$$\int{\left(4 \cos^{2}{\left(x \right)} - 1\right)d x} = x + \sin{\left(2 x \right)}$$
加上積分常數：
$$\int{\left(4 \cos^{2}{\left(x \right)} - 1\right)d x} = x + \sin{\left(2 x \right)}+C$$
答案
$$$\int \left(4 \cos^{2}{\left(x \right)} - 1\right)\, dx = \left(x + \sin{\left(2 x \right)}\right) + C$$$A