JEE Main 2024 — 4 April, Shift 2 — Question 22

$\int \operatorname{cosec}^{3} x \cdot \operatorname{cosec}^{2} x d x=I$

By applying integration by parts

$I=-\cot x \operatorname{cosec}^{3} x+\int \cot x\left(-3 \operatorname{cosec}^{2} x \cot x \operatorname{cosec} x\right) d x$

$I=-\cot x \operatorname{cosec}^{3} x-3 \int \operatorname{cosec}^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x$

$I=-\cot x \operatorname{cosec}^{3} x-3 I+3 \int \operatorname{cosec}^{3} x d x$

let

$I_{1}=\int \operatorname{cosec}^{3} x d x=-\operatorname{cosec} x \cot x-\int \cot ^{2} x \operatorname{cosec} x d x$

$I_{1}=-\operatorname{cosec} x \cot x-\int\left(\operatorname{cosec}^{2} x-1\right) \operatorname{cosec} x d x$

$2 I_{1}=-\operatorname{cosec} x \cot x+\ln \left|\tan \frac{x}{2}\right|$

$I_{1}=-\frac{1}{2} \operatorname{cosec} x \cot x+\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|$

$4 I=-\cot x \operatorname{cosec}^{3} x-\frac{3}{2} \operatorname{cosec} x \cot x+\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+4 c$

$I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^{2} x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+c$

$\therefore \alpha=\frac{-1}{4}, \beta=\frac{3}{8} \rightarrow 8(\alpha+\beta)=1$