find the most general antiderivative or indefinite integral. (hint: 1 + tan²θ = sec²θ) ∫(-3 - 3 tan²θ) dθ…

Explanation:

Step1: Rewrite the integrand

Use the hint $1+\tan^{2}\theta=\sec^{2}\theta$, so $\tan^{2}\theta=\sec^{2}\theta - 1$. Then $- 3-3\tan^{2}\theta=-3-3(\sec^{2}\theta - 1)=-3\sec^{2}\theta$.

Step2: Integrate term - by - term

We know that $\int - 3\sec^{2}\theta d\theta$. Since $\int\sec^{2}\theta d\theta=\tan\theta + C$ (where $C$ is the constant of integration), then $\int - 3\sec^{2}\theta d\theta=-3\tan\theta + C$.

Answer:

$-3\tan\theta + C$