$\int cot (4 x + 9) d x$ is equal to
(where $C$ is the constant of integration)

- \frac{ln | sin (4 x + 9) |}{9} + C

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\frac{ln | sin (4 x + 9) |}{9} + C

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\frac{ln | sin (4 x + 9) |}{4} + C

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- \frac{ln | sin (4 x + 9) |}{4} + C

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Solution

The correct option is C $\frac{ln | sin (4 x + 9) |}{4} + C$
$\int cot (4 x + 9) d x = \frac{ln | sin (4 x + 9) |}{4} + C$
$(∵ cot (a x + b) d x = \frac{ln | sin (a x + b) |}{a} + C, a \neq 0)$

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