Blog

Pattern - 28 of Indefinite Integration

No Comments

⋆ Pattern – 28

= ∫ $\cot^n{x} dx$ , then

$I_n=\ \frac{-\cot^{n-1\ }{x}}{x-1}-I_{n-2}\ :n\geq2$

Here,

$I_n=\ \int\cot^n{x}\ dx$

$=\int\cot^{n-2}{x}\times \cot^2{x}\ dx$

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

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

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

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

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

$\therefore\ \ I_5=\ \frac{-\cot^4{x}}{4}-I_3$

$=\ \frac{-\cot^4{x}}{4}-\left[\frac{-\cot^2{x}}{2}-I_1\right]$

$=\ \frac{-\cot^4{x}}{4}+\frac{\cot^2{x}}{2}+\ cot\int{x}\ dx$

$=\ \frac{-\cot^4{x}}{4}+\frac{\cot^2{x}}{2}+\log{\left|\sin{x}\right|}+c$