Question

1. - / 1.07 points evaluate the integral. (use c for the constant of integration.) ∫ 8 tan(x) sec³(x) dx

Explanation:

Step1: Use substitution

Let $u = \sec(x)$, then $du=\sec(x)\tan(x)dx$. The integral $\int8\tan(x)\sec^{3}(x)dx$ can be rewritten as $8\int\sec^{2}(x)\cdot\sec(x)\tan(x)dx$. Since $u = \sec(x)$, $\sec^{2}(x)=u^{2}$ and $\sec(x)\tan(x)dx = du$. So the integral becomes $8\int u^{2}du$.

Step2: Integrate $u^{2}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), for $n = 2$, we have $\int u^{2}du=\frac{u^{3}}{3}+C$.

Step3: Substitute back $u=\sec(x)$

Substituting $u = \sec(x)$ back into the result, we get $8\times\frac{\sec^{3}(x)}{3}+C=\frac{8}{3}\sec^{3}(x)+C$.

Answer:

$\frac{8}{3}\sec^{3}(x)+C$