$ \int \operatorname{cosec}^3 x \cdot \operatorname{cosec}^2 x d x=1 $
By applying integration by parts
$ \begin{aligned} & I=-\cot x \operatorname{cosec}^3 x+\int \cot x\left(-3 \operatorname{cosec}^2 x \cot x \operatorname{cosec} x\right) d x \\ & I=-\cot x \operatorname{cosec}^3 x-3 \int \operatorname{cosec}^3 x\left(\operatorname{cosec}^2 x-1\right) d x \\ & I=-\cot x \operatorname{cosec}^3 x-3 I+3 \int \operatorname{cosec}^3 x d x \\ & \text { let } \\ & I_1=\int \operatorname{cosec}^3 x d x=-\operatorname{cosec} x \cot x-\int \cot ^2 x \operatorname{cosec} x d x \end{aligned} $
$\begin{aligned} & I_1=-\operatorname{cosec} x \cot x-\int\left(\operatorname{cosec}^2 x-1\right) \operatorname{cosec} x d x 2 I_1=-\operatorname{cosec} x \cot x+\ell n\left|\tan \frac{x}{2}\right| \\ & I_1=-\frac{1}{2} \operatorname{cosec} x \cot x+\frac{1}{2} \ln \left|\tan \frac{x}{2}\right| \\ & 4 I=-\cot x \operatorname{cosec}^3 x-\frac{3}{2} \operatorname{cosec} x \cot x+\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+4 c \\ & I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+c \\ & \therefore \alpha=\frac{-1}{4}, \beta=\frac{3}{8} \rightarrow 8(\alpha+\beta)=1\end{aligned}$