If int operatorname{cosec}^5 x , dx = alpha c | vedclass

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(B) Let $I = \int \operatorname{cosec}^5 x \, dx$. Using integration by parts with $u = \operatorname{cosec}^3 x$ and $dv = \operatorname{cosec}^2 x \

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$1$

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(B) Let $I = \int \operatorname{cosec}^5 x \, dx$. Using integration by parts with $u = \operatorname{cosec}^3 x$ and $dv = \operatorname{cosec}^2 x \, dx$:$I = \operatorname{cosec}^3 x(-\cot x) - \int (3 \operatorname{cosec}^2 x(-\operatorname{cosec} x \cot x))(-\cot x) \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x \cot^2 x \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x (\operatorname{cosec}^2 x - 1) \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3I + 3 \int \operatorname{cosec}^3 x \, dx$$4I = -\operatorname{cosec}^3 x \cot x + 3 \int \operatorname{cosec}^3 x \, dx$We know $\int \operatorname{cosec}^3 x \, dx = -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}|$.Substituting this into the equation for $4I$:$4I = -\operatorname{cosec}^3 x \cot x + 3 \left( -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}| ight)$$4I = -\operatorname{cosec}^3 x \cot x - \frac{3}{2} \operatorname{cosec} x \cot x + \frac{3}{2} \ln |	an \frac{x}{2}|$$I = -\frac{1}{4} \cot x \operatorname{cosec} x (\operatorname{cosec}^2 x + \frac{3}{2}) + \frac{3}{8} \ln |	an \frac{x}{2}| + C$Comparing with the given form,$\alpha = -\frac{1}{4}$ and $\beta = \frac{3}{8}$.Thus,$8(\alpha + \beta) = 8(-\frac{1}{4} + \frac{3}{8}) = 8(\frac{-2+3}{8}) = 1$.

If $\int \operatorname{cosec}^5 x \, dx = \alpha \cot x \operatorname{cosec} x \left(\operatorname{cosec}^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$,where $\alpha, \beta \in R$ and $C$ is the constant of integration,then the value of $8(\alpha + \beta)$ is equal to:

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