# Determine the expression of antiderivative of f(x)=1/cos^4(x)

justaguide | Certified Educator

The question states that f(x) = 1/ (cos x)^4.

We have to find the integral of f(x) = 1/ (cos x)^4.

(cos x)^-4 = (sec x)^4 = (sec x)^2 * (sec x)^2

=> ((tan x)^2 + 1)* (sec x)^2

=> ((tan x)^2*(sec x)^2+ (sec x)^2)

Int [ ((tan x)^2*(sec x)^2+ (sec x)^2)]

=> Int [ ((tan x)^2*(sec x)^2] + Int [ (sec x)^2)]

We know Int [ (sec x)^2)] = tan x.

Now to determine Int [ ((tan x)^2*(sec x)^2]

let  t = tan x , dt / dx = (sec x)^2

=> (sec x)^2 dx = dt

So Int [ ((tan x)^2*(sec x)^2 dx]

=> Int [ t^2 dt]

=> t^3 / 3 + C

=> (tan x )^3 / 3 + C

So Int [ ((tan x)^2*(sec x)^2] + Int [ (sec x)^2)]

=> tan x + (tan x )^3 / 3 + C

The required result is tan x + (tan x )^3 / 3 + C

hala718 | Certified Educator

f(x) = 1/cos^4 x

We need to find the integral of f(x).

intg f(x) = intg (1/cos^4 x ) dx

Let us rewrite :

==> intg f(x) = intg ( 1/cos^2 x* cos^2 x) dx

But we know that 1/cos^2 x= sec^2 x

We also know that sec^2 x = 1+ tan^2 x

==> 1/cos^2 x = 1+ tan^2 x

==> intg f(x) = intg ( 1+tan^2 x)/ cos^2 x   dx

Now we will assume that:

u= tanx ==> du = sec^2 x dx = 1/cos^2 x  dx

==> intg f(x) = intg (1+t^2) * du

=  t + t^3/3 + C

Now we will substitute with t= tanx

==> intg f(x) = tanx + tan^3 x /3  + C

==> intg f(x) = (tan^3 x)/3  + tanx + C

neela | Student

To find Integral of  f(x) = 1/cos^4 (x)

We know that 1/cos^4x. = sec^2x*sec^2x = (1+tan^2x)sec^2x .

Therefore Int f(x) dx =  Int 1/cos^4xdx = (1+tan^2x) sec^2x dx.

Int (1+tan^2x )sec^2x dx = Int sec^2x dx + Int (tan^2x) sec^2 dx.

=  tanx +  Int  t^2 *dt, where  t = tanx and  dt se^2x dx.

= tanx + (t^3)/3 + C.

We replace t = tanx.

Therefore (Int 1/cos^4x )dx = tanx + (1/3)tan^3x +C.

giorgiana1976 | Student

To determine the expression of the antiderivative, we'll have to determine the indefinite integral of the given function.

Int dx/(cos x)^4

We'll re-write the fraction 1/(cos x)^4 = [1/(cos x)^2]*[1/(cos x)^2]

But 1/(cos x)^2 = 1 + (tan x)^2

We'll re-write the integral:

Int [1 + (tan x)^2]*[dx/(cos x)^2]

We'll put tan x = t. We'll differentiate both sides and we'll get:

dx/(cos x)^2 = dt

We'll re-write the integral in t:

Int (t^2 + 1)dt = Int t^2 dt + Int dt

Int (t^2 + 1)dt = t^3/3 + t + C

The antiderivative of the function f(x) = 1/(cos x)^4 is:

Int dx/(cos x)^4 = (tan)^3/3 + tan x + C