Integral of $$$\frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}}$$$ with respect to $$$\pi$$$

Find $$$\int \frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}}\, d\pi$$$.

Apply the constant multiple rule $$$\int c f{\left(\pi \right)}\, d\pi = c \int f{\left(\pi \right)}\, d\pi$$$ with $$$c=\sin^{2}{\left(z \right)}$$$ and $$$f{\left(\pi \right)} = \frac{1}{\left(- \frac{\pi}{6} + z\right)^{3}}$$$:

$${\color{red}{\int{\frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}} d \pi}}} = {\color{red}{\sin^{2}{\left(z \right)} \int{\frac{1}{\left(- \frac{\pi}{6} + z\right)^{3}} d \pi}}}$$

Let $$$u=- \frac{\pi}{6} + z$$$.

Then $$$du=\left(- \frac{\pi}{6} + z\right)^{\prime }d\pi = - \frac{d\pi}{6}$$$ (steps can be seen »), and we have that $$$d\pi = - 6 du$$$.

Therefore,

$$\sin^{2}{\left(z \right)} {\color{red}{\int{\frac{1}{\left(- \frac{\pi}{6} + z\right)^{3}} d \pi}}} = \sin^{2}{\left(z \right)} {\color{red}{\int{\left(- \frac{6}{u^{3}}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-6$$$ and $$$f{\left(u \right)} = \frac{1}{u^{3}}$$$:

$$\sin^{2}{\left(z \right)} {\color{red}{\int{\left(- \frac{6}{u^{3}}\right)d u}}} = \sin^{2}{\left(z \right)} {\color{red}{\left(- 6 \int{\frac{1}{u^{3}} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$$- 6 \sin^{2}{\left(z \right)} {\color{red}{\int{\frac{1}{u^{3}} d u}}}=- 6 \sin^{2}{\left(z \right)} {\color{red}{\int{u^{-3} d u}}}=- 6 \sin^{2}{\left(z \right)} {\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}=- 6 \sin^{2}{\left(z \right)} {\color{red}{\left(- \frac{u^{-2}}{2}\right)}}=- 6 \sin^{2}{\left(z \right)} {\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$

Recall that $$$u=- \frac{\pi}{6} + z$$$:

$$3 \sin^{2}{\left(z \right)} {\color{red}{u}}^{-2} = 3 \sin^{2}{\left(z \right)} {\color{red}{\left(- \frac{\pi}{6} + z\right)}}^{-2}$$

Therefore,

$$\int{\frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}} d \pi} = \frac{3 \sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{2}}$$

Simplify:

$$\int{\frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}} d \pi} = \frac{108 \sin^{2}{\left(z \right)}}{\left(- \pi + 6 z\right)^{2}}$$

Add the constant of integration:

$$\int{\frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}} d \pi} = \frac{108 \sin^{2}{\left(z \right)}}{\left(- \pi + 6 z\right)^{2}}+C$$

$$$\int \frac{\sin^{2}{\left(z \right)}}{\left(- \frac{\pi}{6} + z\right)^{3}}\, d\pi = \frac{108 \sin^{2}{\left(z \right)}}{\left(- \pi + 6 z\right)^{2}} + C$$$A