Mathematics - Indefinite Integration Question with Solution | TestHub

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