$\\int 18x\\cos(9x^{2}+4)dx = \\square$

Explanation:

Step1: Set substitution

Let $u = 9x^{2}+4$, then $du=(18x)dx$.

Step2: Rewrite integral

The integral $\int 18x\cos(9x^{2}+4)dx$ becomes $\int\cos(u)du$.

Step3: Integrate

The antiderivative of $\cos(u)$ is $\sin(u)+C$.

Step4: Substitute back

Substitute $u = 9x^{2}+4$ back, we get $\sin(9x^{2}+4)+C$.

Answer:

$\sin(9x^{2}+4)+C$