Find the integral integrate of (sec(x))^4 dx

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember that `sec x = 1/cos x` , hence `sec^4 x = 1/(cos^4 x)`  such that:

`int sec^4 x dx= int 1/(cos^4 x) dx`

You should use the fundamental formula of trigonometry such that:

`sin^2 x + cos^2 x = 1`

Substituting `sin^2 x + cos^2 x`  for 1 yields:

`int (sin^2 x + cos^2 x)/(cos^4 x) dx`  

Using the property of linearity of integrals yields:

`int (sin^2 x + cos^2 x)/(cos^4 x) dx = int (sin^2 x)/(cos^4 x) dx ` `+ int (cos^2 x)/(cos^4 x) dx `  

`int (sin^2 x + cos^2 x)/(cos^4 x) dx = int tan^2 x*sec^2 x dx + int sec^2 x dx`

You need to solve the integral `int tan^2 x*sec^2 x dx`  using substitution `tan x = t => sec^2 x dx = dt`  such that:

`int tan^2 x*sec^2 x dx = int t^2 dt = t^3/3 + c`

Substituting back `tan x ` for t yields:

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

`int (sin^2 x + cos^2 x)/(cos^4 x) dx = (tan^3 x)/3 + tan x + c`

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

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