If $\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cossc}^2 x+\frac{3}{2} ight)+\beta \log _\epsilon\left| an \frac{x}{2} ight|+C$ where $\alpha, \beta \in \mathbb{R}$ and $\mathrm{C}$ is the constant of integration, then the value of $8(\alpha+\beta)$…

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If $\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cossc}^2 x+\frac{3}{2}ight)+\beta \log _\epsilon\left|	an \frac{x}{2}ight|+C$ where $\alpha, \beta \in \mathbb{R}$ and $\mathrm{C}$ is the constant of integration, then the value of $8(\alpha+\beta)$ equals _______

Aicharya · Accepted Answer

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

If $\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cossc}^2…

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