If \(\int \operatorname{cosec}^{5}x\,dx=\alpha\cot x\,\operatorname{cosec}x\left(\operatorname{cosec}^{2}x+\dfrac{3}{2}\right)+\beta\log_{e}\left|\tan\dfrac{x}{2}\right|+C\) where \(\alpha,\beta\in\mathbb{R}\) and \(C\) is the constant of integration, then the value of \(8(\alpha+\beta)\) equals ____.

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Indefinite Integrals

\int cosec^{5} x d x = α cot x cosec x (cosec^{2} x + \frac{3}{2}) + β lo g_{e} tan \frac{x}{2} + C

where

α, β \in R

and

C

is the constant of integration, then the value of

8 (α + β)

equals ____.

Answer: 1

Reduction formula:

I_{n} = \int csc^{n} x d x = - \frac{csc ^{n - 2} x cot x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2}

I_{1} = \int csc x d x = ln tan \frac{x}{2}

I_{3} = - \frac{csc x cot x}{2} + \frac{1}{2} I_{1} = - \frac{csc x cot x}{2} + \frac{1}{2} ln tan \frac{x}{2}

I_{5} = - \frac{csc ^{3} x cot x}{4} + \frac{3}{4} I_{3} = - \frac{csc ^{3} x cot x}{4} - \frac{3}{8} csc x cot x + \frac{3}{8} ln tan \frac{x}{2} = - cot x csc x (\frac{csc ^{2} x}{4} + \frac{3}{8}) + \frac{3}{8} ln tan \frac{x}{2}

= cot x csc x (csc^{2} x + \frac{3}{2}) \times (- \frac{1}{4}) + \frac{3}{8} ln tan \frac{x}{2}

Matching to the given form:

α = - \frac{1}{4}, β = \frac{3}{8}

8 (α + β) = 8 (- \frac{1}{4} + \frac{3}{8}) = 8 (\frac{1}{8}) = 1

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer