Evaluate $\int cos (- 2 x + 3) d x$

\frac{- sin (- 2 x + 3)}{2} + C

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\frac{- sec (- 2 x + 3)}{2} + C

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\frac{sin (- 2 x + 3)}{3} + C

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\frac{sec (- 2 x + 3)}{3} + C

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Solution

The correct option is A $\frac{- sin (- 2 x + 3)}{2} + C$
As we know that $\int f (a x + b) d x = \frac{F (a x + b)}{a} + C$ , where $a, b$ and $C$ are constants.
We have here,
$f (x) = cos (- 2 x + 3)$ , where $a = - 2, b = 3$
So, $\int cos (- 2 x + 3) d x = \frac{- sin (- 2 x + 3)}{2} + C$

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