Integral of Sec x
2026-02-28 14:00 Diff
https://brightchamps.com/en-us/math/calculus/integral-of-sec-x

In this method, an improper-looking rational function is broken down into a proper rational function. Sec(x) = 1/cos(x)

∫sec (x) dx =  ∫1 / cos(x)

Multiplying and diving by cos(x)

 ∫sec (x) dx =  ∫cos(x) / (cos2x) dx

In trigonometry identities, cos2x = 1 - sin2x

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

u = sin(x), du = cos(x) dx, substituting the value in

∫cos(x) / (1 - sin2x) can be written as ∫du / (1 - u)2

So, ∫sec (x) dx =  ∫du / (1 - u)2

Using partial fraction decomposition on 1 / (1 - u2)

That is, 1 / 1 - u2 = A / (1 + u) + B / (1 - u) 

1 = A(1 - u) +B(1 +u)

1 = A - Au + B + Bu

1 = (A + B) + (-A + B)u

That is A + B = 1

-A + B = 0 → B - A = 0 → B = A

Substituting, B = A into A + B = 1

A + A = 1 →  A = ½, and since B = A, B = ½

Therefore, 1 / 1 - u2 = (1/2)/1 + u + (1/2)/1 - u 

 ∫1 / 1 - u2 du = ½  ∫1 / 1 + u du + ½  ∫ 1 / 1 - u du

 ∫1 / 1 + u = In| 1 + u| and  ∫1 / 1 - u = -In |1 - u| 

So,  ∫1 / 1 - u2 du = ½  In| 1 + u|  + ½  In| 1 - u| + C
∫1 / 1 - u2 du = ½ In |1 + u/1 - u| + C

As u = sin(x), so substituting u = sin(x)

∫ sec(x) dx = ½ In |1 + sin(x)/1 - sin(x)| + C

Therefore, ∫ sec(x) dx = ½ In | 1 + sin(x) / 1 - sin(x) | + C