QJEEM24Ma2982: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2982 Solution: If where and is the constant of integration, then the value of…

$\begin{aligned}& I=\int(\operatorname{cosec} x)^5 d x=\int(\operatorname{cosec} x)^3(\operatorname{cosec} x)^2 d x \\& =(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x- \\& \int\left(\frac{d}{d x}(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x\right) d x\end{aligned}$

$\begin{aligned}& =-\cot x(\operatorname{cosec} x)^3-\int 3 \operatorname{cosec}^2 x \cdot(-\operatorname{cosec} x \cot x)(-\cot x) d x \\& =-\cot x(\operatorname{cosec} x)^3-\int 3 \operatorname{cosec}^3 x \cot ^2 x d x \\& =-\cot x(\operatorname{cosec} x)^3-3 \int(\operatorname{cosec} x)^3\left(\operatorname{cosec}^2 x-1\right) d x \\& I=-\cot x(\operatorname{cosec} x)^3-3 I+3 \int(\operatorname{cosec} x)^3 d x \\& 4 I=-\cot x(\operatorname{cosec} x)^3+3 \int(\operatorname{cosec} x)^3 d x \\& =-\cot x(\operatorname{cosec} x)^3+3 I_1 \\& I_1=\int \operatorname{cosec} x \cdot \operatorname{cosec}^2 x d x=\operatorname{cosec} x(-\cot x)- \\& \qquad \int(-\operatorname{cosec} x \cot x)(-\cot x) d x\end{aligned}$

$\begin{aligned}& I_1=-\operatorname{cosec} x \cot x-\int \operatorname{cosec} x\left(\operatorname{cosec}^2 x-1\right) d x \\& I_1=-\operatorname{cosec} x \cot x-I+\int \operatorname{cosec} x d x \\& 2 l_1=-\operatorname{cosec} x \cot x+\ln \left|\tan \frac{x}{2}\right| \\& \Rightarrow \quad 4 l=-\cot x(\operatorname{cosec} x)^3-\frac{3}{2} \operatorname{cosec} x \cot x +\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+C\end{aligned}$

$\begin{aligned}& I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+C \\& \Rightarrow \alpha=-\frac{1}{4}, \beta=\frac{3}{8} \Rightarrow 8(\alpha+\beta)=1\end{aligned}$