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$\int \frac{\cot^{4} y}{\sin^{2} y} d y is equal to$

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$- \frac{1}{5} {(\cot y)}^{5} + C$

$\frac{1}{5} {(\cot y)}^{5} + C$

$5 {(\cot y)}^{5} + C$

$- 5 {(\cot y)}^{5} + C$

answer is A.

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Detailed Solution

$Let I = \int \frac{\cot^{4} y}{\sin^{2} y} d y$

$I = \int {(\cot y)}^{4} \cos e c^{2} y d y$

$Put \cot y = t \Rightarrow {cosec}^{2} y d y$

$\begin{array}{l} I = \int t^{4} (- d t) \\ I = \frac{- t^{5}}{5} + C \\ I = \frac{- 1}{5} (\cot y)^{5} + C \end{array}$

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