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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Indefinite Integration

If ∫ cosec^5 x d x=α x cosec x(cossc^2 x+(3)/(2))+β _ | (x)/(2)|+C...

If ∫cosec⁡5xdx=αcot⁡xcosec⁡x(cossc⁡2x+32)+βlog⁡ϵ∣tan⁡x2∣+C\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cossc}^2 x+\frac{3}{2}\right)+\beta \log _\epsilon\left|\tan \frac{x}{2}\right|+C where α,β∈R\alpha, \beta \in \mathbb{R} and C\mathrm{C} is the constant of integration, then the value of 8(α+β)8(\alpha+\beta) equals _______

Correct answer

0

Step-by-step solution

∫cosec⁡3x⋅cosec⁡2xdx=I\int \operatorname{cosec}^3 x \cdot \operatorname{cosec}^2 x d x=I
By applying integration by parts I=−cot⁡xcosec⁡3x+∫cot⁡x(−3cosec⁡2xcot⁡xcosec⁡x)dxI=−cot⁡xcosec⁡3x−3∫cosec⁡3x(cosec⁡2x−1)dxI=−cot⁡xcosec⁡3x−3I+3∫cosec⁡3xdx\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 I1=∫cosec⁡3xdx=−cosec⁡xcot⁡x−∫cot⁡2xcosec⁡xdxI1=−cosec⁡xcot⁡x−∫(cosec⁡2x−1)cosec⁡xdx\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}2I1=−cosec⁡xcot⁡x+ln⁡∣tan⁡x2∣I1=−12cosec⁡xcot⁡x+12ln⁡∣tan⁡x2∣4I=−cot⁡xcosec⁡3x−32cosec⁡xcot⁡x+32ln⁡∣tan⁡x2∣+4cI=−14cosec⁡xcot⁡x(cosec⁡2x+32)+38ln⁡∣tan⁡x2∣+c∴α=−14,β=38→8(α+β)=1\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}$
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