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Pattern - 30 of Indefinite Integration

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⋆ Pattern – 30

$I_n=\ \int{\rm cosec}^n\ x\ dx, then$

$I_n=\ \frac{-\cot{x}\times cosec^{n-2\ }x}{n-1}+\ \frac{\left(n-2\right)}{\left(n-1\right)}\ I_{n-2}\ :n\geq2$

Here, $I_n=\int{\rm cosec}^nx dx$ ——-(1)

Here,

$I_n=\ \int{\rm cosec}^{n-2}\times cosec^2x\ dx$

$=cosec^{n-2}x \ dx\ \int dx-\ \int \left(\frac{d}{dx}\ cosec^{n-2}x\ \int {cosec^2x}\ dx\right)\ dx$

$= - cosec^{n-2}x\cot{x} - \int (n-2) cosec^{n-3} \times (-cosec x \times cot x)(-cot x)dx$

= $- cosec^{n-2}x \ cot{x} - (n-2)$ ∫ $\ cosec^{n-2} x \times \cot^2{x} dx$

= $-\ cosec^{n-2} x$ ∙ $cot x - (n-2)$ ∫ $cosec^{n-2}x\left(cosec^2x-1\right) dx$

= – $cosec^{n-2}x$ ∙ $cot x - (n-2)$ ∫ $cosec^nx dx$ + (n-2) ∫ $cosec^{n-2}x$ dx

= – $cosec^{n-2}x$ ∙ cot x – (n-2) + (n-2) $I_{n-2}$

Now,

$\therefore\ \ \left[1+\ \left(n-2\right)\right]\ I_n=\ -cosec^{n-2}x\times \cot{x}+\ \left(n-2\right)I_{n-2}$

$\therefore\ \ I_n=\ \frac{-\cot{x}\times cosec^{n-2\ }x}{n-1}+\ \frac{n-2}{n-1}\ I_{n-2\ }$

1) Evaluate : ∫ $cosec^5x$ dx

Solution. Let, $I_5=\ \int{\rm cosec}^5\ x\ dx$

Here,

$I_5=\ \frac{-\cot{x}\times cosec^3x}{4}+\frac{3}{4}\ I_3$

$=\ \frac{-\cot{x}\times cosec^3x}{4}+\frac{3}{4}\left[\frac{-\cot{x}\times cosec\ x}{2}+\frac{1}{2}\ I_1\right]$

$=\ \frac{-\cot{x}\times cosec^3x}{4}-\frac{3}{8}\cot{x\times }cosec\ x+\frac{3}{8}\ \int{cosec\ x}\ dx$

$=\ \ \frac{-\cot{x}\times cosec^3x}{4}-\frac{3}{8}\cot{x\times }cosec\ x+\frac{3}{8}\ log\left|cosec\ x-\cot{x}\right|+c$