If ∫ cosec^5xdx = α cotx cosecx(cosec^2x + 3/2) + βloge∣tanx/2∣ + C where

Question

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

Answer 1

Correct answer: 1

\(\int \operatorname{cosec}^{3} x \cdot \operatorname{cosec}^{2} x d x=I\)

By applying integration by parts

\(I=-\cot x \operatorname{cosec}^{3} x+\int \cot x\left(-3 \operatorname{cosec}^{2} x \cot x \operatorname{cosec} x\right) d x\)

\(I=-\cot x \operatorname{cosec}^{3} x-3 \int \operatorname{cosec}^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x\)

\(I=-\cot x \operatorname{cosec}^{3} x-3 I+3 \int \operatorname{cosec}^{3} x d x\)

let

\(I_{1}=\int \operatorname{cosec}^{3} x d x=-\operatorname{cosec} x \cot x-\int \cot ^{2} x \operatorname{cosec} x d x\)

\(I_{1}=-\operatorname{cosec} x \cot x-\int\left(\operatorname{cosec}^{2} x-1\right) \operatorname{cosec} x d x\)

\(2 I_{1}=-\operatorname{cosec} x \cot x+\ln\left|\tan \frac{x}{2}\right|\)

\(I_{1}=-\frac{1}{2} \operatorname{cosec} x \cot x+\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|\)

\(4 I=-\cot x \operatorname{cosec}^{3} x-\frac{3}{2} \operatorname{cosec} x \cot x+\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+4 c\)

\(I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^{2} x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+c\)

\(\therefore \alpha=\frac{-1}{4}, \beta=\frac{3}{8} \)

\(\Rightarrow 8(\alpha+\beta)=1\)