Question

\[\int \cot^5 x\ dx\]

Answer 1

\[\text{ Let  I } = \int \cot^5 \text{ x  dx }\]
\[ = \int \cot^2 x \cdot \cot^3\text{ x  dx }\]
\[ = \int\left(\text{cosec}^2 x - 1 \right) \cot^3 \text{ x  dx }\]
\[ = \int \cot^3 x \cdot \text{cosec}^2\text{ x  dx } - \int \cot^3 \text{ x  dx }\]
\[ = \int \cot^3 x \cdot \text{cosec}^2 \text{ x  dx }- \int\cot x \cdot \cot^2 \text{ x  dx }\]
\[ = \int \cot^3 x \cdot \text{cosec}^2\text{ x  dx }- \int\cot x \left( {cosec}^2 x - 1 \right)\text{   dx }\]
\[ = \int \cot^3 x \cdot \text{cosec}^2 \text{ x  dx } - \int\cot x \cdot \text{ cosec}^2\text{ x  dx }+ \int\cot\text{ x  dx }\]
\[\text{   Putting cot  x  = t   in  the  Ist  and  IInd  integral}\]
\[ \Rightarrow - \text{cosec}^2\text{ x  dx }= dt\]
\[ \Rightarrow \text{cosec}^2 \text{ x  dx }= - dt\]
\[ \therefore I = - \int t^3 dt + \int t \cdot dt + \int\cot\text{ x  dx }\]
\[ = - \frac{t^4}{4} + \frac{t^2}{2} + \text{ ln }\left| \sin x \right| + C\]
\[ = - \frac{\cot^4 x}{4} + \frac{\cot^2 x}{2} + \text{ ln }\left| \sin x \right| + C .........\left( \because t = \cot x \right)\]