Integral of Sec x

To find the integral of sec x, we will have to use some facts from trigonometry. Sec x is the reciprocal of cos x and tan x can be written as (sin x)/(cos x). We can do the integration of secant x in multiple methods such as:

• By using substitution method
• By using partial fractions
• By using trigonometric formulas
• By using hyperbolic functions

We have multiple formulas for integration of sec x and let us derive each of them using the above mentioned methods. Also, we will solve some examples related to the integral of sec x.

The integral of sec x is ln|sec x + tan x| + C. It denoted by ∫ sec x dx. This is also known as the antiderivative of sec x. We have multiple formulas for this. But the more popular formula is, ∫ sec x dx = ln |sec x + tan x| + C. Here "ln" stands for natural logarithm and 'C' is the integration constant. Multiple formulas for the integral of sec x are listed below:

• ∫ sec x dx = ln |sec x + tan x| + C [OR]
• ∫ sec x dx = (1/2) ln | (1 + sin x) / (1 - sin x) | + C [OR]
• ∫ sec x dx = ln | tan [ (x/2) + (π/4) ] | + C
• ∫ sec x dx = cosh^-1(sec x) + C (or) sinh^-1(tan x) + C (or) tanh^-1(sin x) + C

We use one of these formulas according to necessity.

We will prove each of these formulas in different methods. Sounds interesting? Let's go!

We can find the integral of sec x by substitution method. For this, we multiply and divide the integrand with (sec x + tan x). But why do we need to do this? Let us see.

∫ sec x dx = ∫ sec x · (sec x + tan x) / (sec x + tan x) dx

= ∫ (sec²x + sec x tan x) / (sec x + tan x) dx

Now assume that sec x + tan x = u.

Then (sec x tan x + sec²x) dx = du.

Substituting these values in the above integral,

∫ sec x dx = ∫ du / u = ln |u| + C

Substituting u = sec x + tan x back here,

∫ sec x dx = ln |sec x + tan x| + C

Hence proved.

To find the integration of sec x by partial fractions, we have to use the fact that sec x is the reciprocal of cos x. Wait! How can this be turned into partial fractions? Let us see.

∫ sec x dx = ∫ 1/(cos x) dx

Multiplying and dividing this by cos x,

∫ sec x dx = ∫ (cos x) / (cos²x) dx

Using one of the trigonometric identities,

∫ sec x dx = ∫ (cos x dx) / (1 - sin²x)

Now, assume that sin x = u. Then cos x dx = du. Then the above integral becomes

∫ sec x dx = ∫ du / (1 - u²)

By partial fractions, 1/(1 - u²) = 1/2 [1/(1 + u) + 1/(1 - u)]. Then

∫ sec x dx = (1/2) ∫ [ 1/(1 + u) + 1/(1 - u) ] du

= (1/2) [ ln |1 + u| - ln |1 - u| ] + C

(We can get this by finding the integrals ∫ [ 1/(1 + u) ] du and ∫ [ 1/(1 - u) ] du separately by substitutiong 1 + u = p and 1 - u = q respectively).

By a property of logarithms, ln m - ln n = ln (m/n). So

∫ sec x dx = (1/2) ln | (1 + u) / (1 - u) | + C

Substituting u = sin x back here,

∫ sec x dx = (1/2) ln | (1 + sin x) / (1 - sin x) | + C

Hence proved.

We can prove that the integral of sec x to be ln | tan [ (x/2) + (π/4) ] | + C by using trigonometric formulas. We can write sec x as 1/(cos x) where cos x can again be written as sin(x + π/2). So

∫ sec x dx = ∫ 1/(cos x) dx

= ∫ 1/(sin(x + π/2)) dx

Using half-angle formulas, sin A = 2 sin A/2 cos A/2. Applying this,

∫ sec x dx = ∫ 1 / [ 2 sin[ (x/2) + (π/4) ] cos[ (x/2) + (π/4) ] ] dx

= (1/2) ∫ 1 / [ sin[ (x/2) + (π/4) ] cos[ (x/2) + (π/4) ] ] dx

Multiplying and dividing the denominator by cos[ (x/2) + (π/4) ],

∫ sec x dx = (1/2) ∫ 1 / [ sin[ (x/2) + (π/4) ]/cos[ (x/2) + (π/4) ] · cos²[ (x/2) + (π/4) ] ] dx

= (1/2) ∫ sec²[ (x/2) + (π/4) ] / tan[ (x/2) + (π/4) ] dx

Now, assume that tan[(x/2) + (π/4)] = u. From this, (1/2) sec²[ (x/2) + (π/4) ] dx = du.

∫ sec x dx = ∫ du/u = ln |u| + C

Substituting u = tan[(x/2) + (π/4)] back,

∫ sec x dx = ln | tan [ (x/2) + (π/4) ] | + C

Hence proved.

Using hyperbolic functions, we can prove that the integral of sin x to be tanh^-1 (sin x) + C. For this, we use the substitution sec x = cosh t. Now, we have

tan x = √sec²x - 1 = √cosh²t - 1 = √sinh²t = sinh t

Differentiating both sides,

sec²x dx = cosh t dt

But sec x = cosh t.

(cosh²t) dx = cosh t dt

dx = (cosh t) / (cosh²t) dt = 1/(cosh t) dt

Substituting these values in ∫ sec x dx,

∫ sec x dx = ∫ (cosh t) [1/(cosh t) dt]

= ∫ dt

= t

= cosh^-1(sec x) + C (This is because sec x = cosh t)

Similarly we can prove that ∫ sec x dx = sinh^-1(tan x) + C (or) ∫ sec x dx = tanh^-1(sin x) + C. Can you give them a try?

Therefore, ∫ sec x dx = cosh^-1(sec x) + C (or) sinh^-1(tan x) + C (or) tanh^-1(sin x) + C.

Hence proved.

Important Notes Related to Integration of Sec x:

Here are the formulas of integral of secant x with the respective methods of proving them.

• Using the substitution method,
  ∫ sec x dx = ln |sec x + tan x| + C
• Using partial fractions,
  ∫ sec x dx = (1/2) ln | (1 + sin x) / (1 - sin x) | + C
• Using trigonometric formulas,
  ∫ sec x dx = ln | tan [ (x/2) + (π/4) ] | + C
• Using hyperbolic functions,
  ∫ sec x dx = cosh^-1(sec x) + C (or) sinh^-1(tan x) + C (or) tanh^-1(sin x) + C

Topics Related to Integral of Sec x:

Here are some topics that you may find helpful while doing the integration of sec x.

• Integration Formulas
• Integral Calculator
• Derivative Formulas
• Derivative Calculator

What is the Integral of Sec x?

There are multiple formulas for the integral of sec x. But the most popular formula that we use is ∫ sec x dx = ln |sec x + tan x| + C, where C is the integration constant.

How do You Find the Antiderivative of Sec x?

The antiderivative of sec x is mathematically writen as ∫ sec x dx. Multiply and divide sec x by (sec x + tan x), we get, ∫ sec x dx = ∫ sec x · (sec x + tan x) / (sec x + tan x) dx = ∫ (sec²x + sec x tan x) / (sec x + tan x) dx. By assuming sec x + tan x = u. Then (sec x tan x + sec²x) dx = du. Then the above integral becomes ∫ sec x dx = ∫ du / u = ln |u| + C = ln |sec x + tan x| + C.

What is the Integral of Sec x Cos x?

We know that sec x and cos x are reciprocals of each other. So sec x cos x = 1. Thus, ∫ sec x cos x dx = ∫ dx = x + C.

What is the Integration of Sec x Tan x?

We know that the derivative of sec x is sec x tan x. Thus, the integral of sec x tan x is sec x. i.e., ∫ sec x tan x dx = sec x + C.

How to Find the Integral of Sec x by Substitution?

We can write ∫ sec x dx as ∫ sec x dx = ∫ sec x · (sec x + tan x) / (sec x + tan x) dx. Now assume sec x + tan x = u. Differentiating, (sec x tan x + sec²x) dx = du. Then, ∫ sec x dx = ∫ du / u = ln |u| + C = ln |sec x + tan x| + C.

What is the Integral of Sec²x Tan x?

We can write ∫ sec²x tan x dx as ∫ sec x (sec x tan x) dx. Assume that sec x = u then sec x tan x dx = du. Then the above integral becomes, ∫ u du = u²/2 + C = (sec²x)/2 + C.

What is the Integral of Sec x from 0 to π/4?

We know that the integral of secant x is ln |sec x + tan x|. Applying the limits 0 and π/4, we get ∫₀^π/4 sec x dx = ln |sec π/4 + tan π/4| - ln |sec 0 + tan 0| = ln |√2 + 1| - ln 1 = ln |√2 + 1|.