Answer
$\dfrac{-1}{6} \sin (-3x^2+4 )+C$
Work Step by Step
Our aim is to solve the integral $\int x \cos (-3x^2+4 ) \ dx$
Let us consider that $a =-3x^2+4$ and $\dfrac{da}{dx}=-6x\implies dx=\dfrac{-da}{6x}$
Now, $\int x \cos (-3x^2+4 ) \ dx = \int x \cos a \times \dfrac{-1}{6x} \ da$
or, $=\dfrac{-1}{6} \sin a+C$
or, $=\dfrac{-1}{6} \sin (-3x^2+4 )+C$
