$$$4 \cos^{4}{\left(\frac{\theta}{2} \right)}$$$ 的積分

求$$$\int 4 \cos^{4}{\left(\frac{\theta}{2} \right)}\, d\theta$$$。

套用常數倍法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$，使用 $$$c=4$$$ 與 $$$f{\left(\theta \right)} = \cos^{4}{\left(\frac{\theta}{2} \right)}$$$：

$${\color{red}{\int{4 \cos^{4}{\left(\frac{\theta}{2} \right)} d \theta}}} = {\color{red}{\left(4 \int{\cos^{4}{\left(\frac{\theta}{2} \right)} d \theta}\right)}}$$

令 $$$u=\frac{\theta}{2}$$$。

則 $$$du=\left(\frac{\theta}{2}\right)^{\prime }d\theta = \frac{d\theta}{2}$$$ (步驟見»)，並可得 $$$d\theta = 2 du$$$。

該積分變為

$$4 {\color{red}{\int{\cos^{4}{\left(\frac{\theta}{2} \right)} d \theta}}} = 4 {\color{red}{\int{2 \cos^{4}{\left(u \right)} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=2$$$ 與 $$$f{\left(u \right)} = \cos^{4}{\left(u \right)}$$$：

$$4 {\color{red}{\int{2 \cos^{4}{\left(u \right)} d u}}} = 4 {\color{red}{\left(2 \int{\cos^{4}{\left(u \right)} d u}\right)}}$$

套用降冪公式 $$$\cos^{4}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$，令 $$$\alpha= u $$$:

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{8}$$$ 與 $$$f{\left(u \right)} = 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$：

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

逐項積分：

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

配合 $$$c=3$$$，應用常數法則 $$$\int c\, du = c u$$$：

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=4$$$ 與 $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$：

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

令 $$$v=2 u$$$。

則 $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步驟見»)，並可得 $$$du = \frac{dv}{2}$$$。

因此，

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

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$：

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

餘弦函數的積分為 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$：

$$3 u + \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\int{\cos{\left(v \right)} d v}}} = 3 u + \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\sin{\left(v \right)}}}$$

回顧一下 $$$v=2 u$$$：

$$3 u + \int{\cos{\left(4 u \right)} d u} + 2 \sin{\left({\color{red}{v}} \right)} = 3 u + \int{\cos{\left(4 u \right)} d u} + 2 \sin{\left({\color{red}{\left(2 u\right)}} \right)}$$

令 $$$v=4 u$$$。

則 $$$dv=\left(4 u\right)^{\prime }du = 4 du$$$ (步驟見»)，並可得 $$$du = \frac{dv}{4}$$$。

因此，

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

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$：

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

餘弦函數的積分為 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$：

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

回顧一下 $$$v=4 u$$$：

$$3 u + 2 \sin{\left(2 u \right)} + \frac{\sin{\left({\color{red}{v}} \right)}}{4} = 3 u + 2 \sin{\left(2 u \right)} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{4}$$

回顧一下 $$$u=\frac{\theta}{2}$$$：

$$2 \sin{\left(2 {\color{red}{u}} \right)} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{4} + 3 {\color{red}{u}} = 2 \sin{\left(2 {\color{red}{\left(\frac{\theta}{2}\right)}} \right)} + \frac{\sin{\left(4 {\color{red}{\left(\frac{\theta}{2}\right)}} \right)}}{4} + 3 {\color{red}{\left(\frac{\theta}{2}\right)}}$$

因此，

$$\int{4 \cos^{4}{\left(\frac{\theta}{2} \right)} d \theta} = \frac{3 \theta}{2} + 2 \sin{\left(\theta \right)} + \frac{\sin{\left(2 \theta \right)}}{4}$$

化簡：

$$\int{4 \cos^{4}{\left(\frac{\theta}{2} \right)} d \theta} = \frac{6 \theta + 8 \sin{\left(\theta \right)} + \sin{\left(2 \theta \right)}}{4}$$

加上積分常數：

$$\int{4 \cos^{4}{\left(\frac{\theta}{2} \right)} d \theta} = \frac{6 \theta + 8 \sin{\left(\theta \right)} + \sin{\left(2 \theta \right)}}{4}+C$$

$$$\int 4 \cos^{4}{\left(\frac{\theta}{2} \right)}\, d\theta = \frac{6 \theta + 8 \sin{\left(\theta \right)} + \sin{\left(2 \theta \right)}}{4} + C$$$A