Question

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

• A. $\frac{-\sin (-2x+3)}{2}+C$
• B. $\frac{-\sec (-2x+3)}{2}+C$
• C. $\frac{\sin (-2x+3)}{3}+C$
• D. $\frac{\sec (-2x+3)}{3}+C$

Answer 1

The correct option is A $\frac{-\sin (-2x+3)}{2}+C$

As we know that $\int f(ax+b)dx=\frac{F(ax+b)}{a}+C$, where $a,b$ and $C$ are constants.
We have here,
$f\left(x\right)=\cos (-2x+3)$, where $a=-2,b=3$
So, $\int \cos (-2x+3)dx=\frac{-\sin (-2x+3)}{2}+C$