$$$\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}$$$ 對 $$$\pi$$$ 的積分

求$$$\int \frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}\, d\pi$$$。

套用常數倍法則 $$$\int c f{\left(\pi \right)}\, d\pi = c \int f{\left(\pi \right)}\, d\pi$$$，使用 $$$c=\sin^{2}{\left(z \right)}$$$ 與 $$$f{\left(\pi \right)} = \frac{1}{- \frac{\pi}{6} + z}$$$：

$${\color{red}{\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi}}} = {\color{red}{\sin^{2}{\left(z \right)} \int{\frac{1}{- \frac{\pi}{6} + z} d \pi}}}$$

令 $$$u=- \frac{\pi}{6} + z$$$。

則 $$$du=\left(- \frac{\pi}{6} + z\right)^{\prime }d\pi = - \frac{d\pi}{6}$$$ (步驟見»)，並可得 $$$d\pi = - 6 du$$$。

因此，

$$\sin^{2}{\left(z \right)} {\color{red}{\int{\frac{1}{- \frac{\pi}{6} + z} d \pi}}} = \sin^{2}{\left(z \right)} {\color{red}{\int{\left(- \frac{6}{u}\right)d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-6$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$$\sin^{2}{\left(z \right)} {\color{red}{\int{\left(- \frac{6}{u}\right)d u}}} = \sin^{2}{\left(z \right)} {\color{red}{\left(- 6 \int{\frac{1}{u} d u}\right)}}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$- 6 \sin^{2}{\left(z \right)} {\color{red}{\int{\frac{1}{u} d u}}} = - 6 \sin^{2}{\left(z \right)} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回顧一下 $$$u=- \frac{\pi}{6} + z$$$：

$$- 6 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} \sin^{2}{\left(z \right)} = - 6 \ln{\left(\left|{{\color{red}{\left(- \frac{\pi}{6} + z\right)}}}\right| \right)} \sin^{2}{\left(z \right)}$$

因此，

$$\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi} = - 6 \ln{\left(\left|{\frac{\pi}{6} - z}\right| \right)} \sin^{2}{\left(z \right)}$$

化簡：

$$\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi} = 6 \left(- \ln{\left(\left|{\pi - 6 z}\right| \right)} + \ln{\left(6 \right)}\right) \sin^{2}{\left(z \right)}$$

加上積分常數：

$$\int{\frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z} d \pi} = 6 \left(- \ln{\left(\left|{\pi - 6 z}\right| \right)} + \ln{\left(6 \right)}\right) \sin^{2}{\left(z \right)}+C$$

$$$\int \frac{\sin^{2}{\left(z \right)}}{- \frac{\pi}{6} + z}\, d\pi = 6 \left(- \ln\left(\left|{\pi - 6 z}\right|\right) + \ln\left(6\right)\right) \sin^{2}{\left(z \right)} + C$$$A