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