AP Calculus AB and BC

Integral of sec 2x: Answer and the Trick

The integral of sec 2x is one half times the natural log of the absolute value of sec 2x plus tan 2x, plus C. It follows from the standard secant integral, with a one half from the inner coefficient. The derivation multiplies by secant plus tangent over itself.

∫sec⁡2x dx=12ln⁡∣sec⁡2x+tan⁡2x∣+C\int \sec 2x\,dx = \frac{1}{2}\ln\left|\sec 2x + \tan 2x\right| + C

Where the answer comes from

The base case uses a trick that looks unmotivated until you see the numerator: multiplying by sec⁡u+tan⁡usec⁡u+tan⁡u\frac{\sec u + \tan u}{\sec u + \tan u} makes the numerator exactly the derivative of the denominator.

ddu(sec⁡u+tan⁡u)=sec⁡utan⁡u+sec⁡2u=sec⁡u(sec⁡u+tan⁡u)\frac{d}{du}\left(\sec u + \tan u\right) = \sec u\tan u + \sec^{2}u = \sec u\left(\sec u + \tan u\right)

So the integral becomes ∫dvv\int \frac{dv}{v} with v=sec⁡u+tan⁡uv = \sec u + \tan u, and the inner 22 contributes the 12\frac{1}{2}.

Worth memorising

This is not a derivation anyone reconstructs under exam pressure. Both secant and cosecant integrals are memorised results on the AP formula sheet, and the derivation is worth seeing once so the answer is not arbitrary.

∫sec⁡u du=ln⁡∣sec⁡u+tan⁡u∣+C\int \sec u\,du = \ln\left|\sec u + \tan u\right| + C

Common mistakes

  • Losing the 12\frac{1}{2} from the inner coefficient.
  • Writing sec⁡2x−tan⁡2x\sec 2x - \tan 2x inside. The standard form adds.
  • Answering ln⁡∣sec⁡2x∣\ln\left|\sec 2x\right|, which is the tangent integral's form instead.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of sec 2x?

It is 12ln⁡∣sec⁡2x+tan⁡2x∣+C\frac{1}{2}\ln\left|\sec 2x + \tan 2x\right| + C.

Do I need to know the derivation?

No. It is a memorised formula sheet result, but seeing the trick once explains why the answer has that shape.