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∫ Cot 4 X D X

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Question

\[\int \cot^4 x\ dx\]
Sum
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Solution

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

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