compute the given integral.\n$intsec^{3}x dx=$

Explanation:

Step1: Use integration - by - parts

Let $u = \sec x$ and $dv=\sec^{2}x dx$. Then $du=\sec x\tan xdx$ and $v = \tan x$. By the integration - by - parts formula $\int u dv=uv-\int v du$, we have $\int\sec^{3}x dx=\sec x\tan x-\int\tan x\cdot\sec x\tan xdx=\sec x\tan x-\int\sec x\tan^{2}x dx$.

Step2: Rewrite $\tan^{2}x$

Since $\tan^{2}x=\sec^{2}x - 1$, then $\int\sec x\tan^{2}x dx=\int\sec x(\sec^{2}x - 1)dx=\int\sec^{3}x dx-\int\sec x dx$.

Step3: Substitute and solve for $\int\sec^{3}x dx$

We had $\int\sec^{3}x dx=\sec x\tan x-(\int\sec^{3}x dx-\int\sec x dx)$. Add $\int\sec^{3}x dx$ to both sides: $2\int\sec^{3}x dx=\sec x\tan x+\int\sec x dx$. We know that $\int\sec x dx=\ln|\sec x+\tan x| + C$. So, $\int\sec^{3}x dx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C$.

Answer:

$\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x + \tan x|+C$