$\int sin (9 x + 24) d x$ is equal to
(where $C$ is constant of integration)

\frac{cos (9 x + 24)}{9} + C

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\frac{- cos (9 x + 24)}{9} + C

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\frac{- cos (9 x + 24)}{24} + C

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\frac{cos (9 x + 24)}{24} + C

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Solution

The correct option is B $\frac{- cos (9 x + 24)}{9} + C$
$\int sin (9 x + 24) d x = \frac{- cos (9 x + 24)}{9} + C$ $(∵ \int sin (a x + b) d x = \frac{- cos (a x + b)}{a} + C, a \neq 0)$

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