12. - / 2.12 points evaluate the integral. (remember the constant of integration.) $int 4\tan^{3}(x)sec(x)dx$

Explanation:

Step1: Rewrite $\tan^{3}(x)$

We know that $\tan^{3}(x)=\tan^{2}(x)\tan(x) = (\sec^{2}(x)- 1)\tan(x)$. So the integral $\int4\tan^{3}(x)\sec(x)dx=4\int(\sec^{2}(x)-1)\tan(x)\sec(x)dx$.

Step2: Use substitution

Let $u = \sec(x)$, then $du=\sec(x)\tan(x)dx$. The integral becomes $4\int(u^{2}-1)du$.

Step3: Integrate term - by - term

Integrating $4\int(u^{2}-1)du$ gives $4\left(\int u^{2}du-\int 1du\right)$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $4\left(\frac{u^{3}}{3}-u\right)+C$.

Step4: Substitute back $u=\sec(x)$

Substituting $u = \sec(x)$ back into the result, we get $4\left(\frac{\sec^{3}(x)}{3}-\sec(x)\right)+C=\frac{4}{3}\sec^{3}(x)-4\sec(x)+C$.

Answer:

$\frac{4}{3}\sec^{3}(x)-4\sec(x)+C$