# Find the integral integrate of (sec(x))^5 dx

*print*Print*list*Cite

### 1 Answer

You should use the following formula such that:

`int sec^k x dx = (sin x*sec^(k-1)x)/(k-1) + (k-2)/(k-1)int sec^(k-2)x dx`

Considering k=5 yields:

`int sec^5 x dx = (sin x*sec^4 x)/4 + 3/4 int sec^3 x dx`

You need to evaluate `int` `sec^3 x` dx using the formula above such that:

`int sec^3 x dx = (sin x*sec^2 x)/2 +(1/2) int sec x dx`

`int sec^3 x dx = (sin x*sec^2 x)/2 + (1/2)log(tan x + sec x) + c`

Substituting `(sin x*sec^2 x)/2 + (1/2)log(tan x + sec x) + c ` for `int sec^3 x dx` yields:

`int sec^5 x dx = (sin x*sec^4 x)/4 + 3/4((sin x*sec^2 x)/2 + (1/2)log(tan x + sec x)) + c`

**Hence, evaluating the given integral yields** **`int sec^5 x dx = (sin x*sec^4 x)/4 + 3/4((sin x*sec^2 x)/2 + (1/2)log(tan x + sec x)) + c.` **