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Trigonometric integrals

Integral of sec³ x: Formula, Derivation and Examples

The integral of sec³ x is ½(sec x tan x + ln|sec x + tan x|) + C. See the integration by parts loop step by step, a full check and the arc length application.

Trigonometric integrals Updated 3 min read

At a glance

\int \sec^3 x\,dx = \frac{1}{2}\left(\sec x\tan x + \ln\lvert\sec x + \tan x\rvert\right) + C
Antiderivative
\dfrac{1}{2}\sec x\tan x + \dfrac{1}{2}\ln\lvert\sec x + \tan x\rvert + C
Method
Integration by parts with u = \sec x, dv = \sec^2 x\,dx
Key identity
\tan^2 x = \sec^2 x - 1
From 0 to π/4
\dfrac{\sqrt{2} + \ln(1 + \sqrt{2})}{2} \approx 1.148
Reduction formula
\int\sec^n x\,dx = \dfrac{\sec^{n-2}x\tan x}{n - 1} + \dfrac{n - 2}{n - 1}\int\sec^{n-2}x\,dx
On this page
  1. Graph of sec³ x and the area under it
  2. Derivation by parts, step by step
  3. Check by differentiating
  4. Definite integral
  5. Application: integral of √(1 + x²) and arc length
  6. The reduction formula
  7. Common mistakes

The integral of \sec^3 x is \dfrac{1}{2}\left(\sec x\tan x + \ln\lvert\sec x + \tan x\rvert\right) + C. It is the classic example of integration by parts where the integral comes back around and you solve for it.

\int \sec^3 x\,dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln\lvert\sec x + \tan x\rvert + C

Graph of sec³ x and the area under it

Area under y = sec³ x from 0 to π/4
The shaded area is \frac{\sqrt{2} + \ln(1 + \sqrt{2})}{2} \approx 1.148. The function grows very quickly toward the asymptote at x = \frac{\pi}{2}.

Derivation by parts, step by step

Call the integral I = \int\sec^3 x\,dx.

  1. Split \sec^3 x = \sec x\cdot\sec^2 x and use the integration by parts rule with u = \sec x,\quad dv = \sec^2 x\,dx,\qquad du = \sec x\tan x\,dx,\quad v = \tan x
  2. Apply \int u\,dv = uv - \int v\,du: I = \sec x\tan x - \int\sec x\tan^2 x\,dx
  3. Replace \tan^2 x with \sec^2 x - 1: I = \sec x\tan x - \int\sec^3 x\,dx + \int\sec x\,dx = \sec x\tan x - I + \int\sec x\,dx
  4. The original integral I has appeared on the right. Add I to both sides: 2I = \sec x\tan x + \ln\lvert\sec x + \tan x\rvert
  5. Divide by 2 and add C.

Step 4 uses the integral of sec x. The choice dv = \sec^2 x\,dx works because \sec^2 x is the one piece you can integrate easily, to \tan x. The integration by parts guide has more examples of this loop.

Check by differentiating

First, by the product rule:

\frac{d}{dx}(\sec x\tan x) = \sec x\tan x\cdot\tan x + \sec x\cdot\sec^2 x = \sec x\tan^2 x + \sec^3 x

Replace \tan^2 x with \sec^2 x - 1 to get 2\sec^3 x - \sec x. The log term differentiates to \sec x. So

\frac{d}{dx}\left[\frac{1}{2}\sec x\tan x + \frac{1}{2}\ln\lvert\sec x + \tan x\rvert\right] = \frac{1}{2}\left(2\sec^3 x - \sec x + \sec x\right) = \sec^3 x \checkmark

Definite integral

At x = \frac{\pi}{4}: \sec = \sqrt{2}, \tan = 1. At x = 0: \sec = 1, \tan = 0, and \ln 1 = 0. So

\int_0^{\pi/4}\sec^3 x\,dx = \frac{1}{2}\left(\sqrt{2} + \ln(\sqrt{2} + 1)\right) \approx 1.148

Application: integral of √(1 + x²) and arc length

Let x = \tan\theta, dx = \sec^2\theta\,d\theta. Then \sqrt{1 + x^2} = \sec\theta and

\int\sqrt{1 + x^2}\,dx = \int\sec^3\theta\,d\theta = \frac{1}{2}x\sqrt{1 + x^2} + \frac{1}{2}\ln\left(x + \sqrt{1 + x^2}\right) + C

The arc length of the parabola y = \frac{x^2}{2} from x = 0 to x = 1 is \int_0^1\sqrt{1 + x^2}\,dx, which is exactly the shaded area above: about 1.148. More on this in the lesson on arc length and surface area.

The reduction formula

The same steps with \sec^n x give

\int\sec^n x\,dx = \frac{\sec^{n-2}x\tan x}{n - 1} + \frac{n - 2}{n - 1}\int\sec^{n-2}x\,dx

With n = 3 this is our result. With n = 5, it reduces \sec^5 x to \sec^3 x, which you now know.

Common mistakes

  1. Using the power rule: \frac{\sec^4 x}{4}. Its derivative is \sec^4 x\tan x, not \sec^3 x.
  2. Choosing u = \sec^2 x, dv = \sec x\,dx. Then v = \ln\lvert\sec x + \tan x\rvert, and the new integral is much harder.
  3. Forgetting to divide by 2 after collecting the I terms.
  4. Losing the log term. The \int\sec x\,dx from step 3 is easy to drop.

Practise on integration by parts problems.

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Frequently asked questions

What is the integral of sec³ x?
It is \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln\lvert\sec x + \tan x\rvert + C.

Why can't I use the power rule on sec³ x?
The power rule for \int u^3\,du needs du = \sec x\tan x\,dx to be in the integral. It is not, so you need integration by parts instead.

What is the integral of √(1 + x²)?
With x = \tan\theta it becomes \int\sec^3\theta\,d\theta. The answer is \frac{1}{2}x\sqrt{1 + x^2} + \frac{1}{2}\ln\left(x + \sqrt{1 + x^2}\right) + C.

What is the integral of sec³ x from 0 to π/4?
It is \frac{1}{2}\left(\sqrt{2} + \ln(1 + \sqrt{2})\right) \approx 1.148.

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Related integrals

Integral of sec x \int \sec x\,dx = \ln\lvert\sec x + \tan x\rvert + C The integral of sec x is ln|sec x + tan x| + C. See the multiply-by-one trick step by step, the equivalent forms, a check by differentiating and examples. Integral of sec² x \int \sec^2 x\,dx = \tan x + C The integral of sec² x is tan x + C. See why it follows from the derivative of tan x, the sec²(ax) form, related integrals like tan² x and csc² x, and examples. Integral of tan² x \int \tan^2 x\,dx = \tan x - x + C The integral of tan² x is tan x - x + C. See how the identity tan² x = sec² x - 1 does all the work, plus tan²(ax), tan³ x and definite integral examples.