evaluate the integral.\n int 8\tan^{3}xmathrm{d}x \n int 8\tan^{3}xmathrm{d}x=square \n(use parentheses to…

Explanation:

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

We know that $\tan^{3}x=\tan x\cdot\tan^{2}x=\tan x(\sec^{2}x - 1)$. So, $\int8\tan^{3}x dx=8\int\tan x(\sec^{2}x - 1)dx=8\left(\int\tan x\sec^{2}x dx-\int\tan x dx\right)$.

Step2: Integrate $\int\tan x\sec^{2}x dx$

Let $u = \tan x$, then $du=\sec^{2}x dx$. So, $\int\tan x\sec^{2}x dx=\int u du=\frac{u^{2}}{2}+C_1=\frac{\tan^{2}x}{2}+C_1$.

Step3: Integrate $\int\tan x dx$

We know that $\int\tan x dx=-\ln|\cos x|+C_2$.

Step4: Combine the results

$8\left(\int\tan x\sec^{2}x dx-\int\tan x dx\right)=8\left(\frac{\tan^{2}x}{2}+\ln|\cos x|\right)+C = 4\tan^{2}x+8\ln|\cos x|+C$.

Answer:

$4\tan^{2}x + 8\ln|\cos x|+C$