Answer
$$\int e^t\cos(3e^t-2)dt=\frac{\sin(3e^t-2)}{3}+C$$
Work Step by Step
$$A=\int e^t\cos(3e^t-2)dt$$
Set $u=3e^t-2$, which means $$du=3e^tdt$$ $$e^tdt=\frac{1}{3}du$$
Therefore, $$A=\frac{1}{3}\int\cos udu$$ $$A=\frac{\sin u}{3}+C$$ $$A=\frac{\sin(3e^t-2)}{3}+C$$
