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

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### 4 Answers

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**

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**

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.

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**