JEE Main 2024 — Indefinite Integration Question with Solution

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