AP Calculus AB and BC

Integral of sec x tan x: Answer and Proof

Q: What is the integral of sec x tan x?

It is \sec x + C. The product \sec x\tan x is the derivative of \sec x, so reversing that derivative gives \sec x.

Q: Does the answer have a minus sign?

No. \int \sec x\tan x\,dx = \sec x + C with no minus. The minus sign belongs to the cosecant pair, \int \csc x\cot x\,dx = -\csc x + C.

Q: How is it different from the integral of sec^2 x?

\int \sec^2 x\,dx = \tan x + C, while \int \sec x\tan x\,dx = \sec x + C. Different integrands come from different standard derivatives.

The integral of sec x tan x with respect to x is sec x + C. The product sec x tan x is exactly the derivative of sec x, so the antiderivative is sec x itself. Differentiating sec x returns sec x tan x, which confirms the answer.

\int \sec x\tan x\,dx = \sec x + C

How to integrate sec x tan x

This integrand is a memorized derivative in disguise. The product $\sec x\tan x$ is not something you attack with substitution or parts; you recognize it as the right side of a standard derivative rule and read the antiderivative backward.

\frac{d}{dx}\sec x = \sec x\tan x

Since differentiating $\sec x$ produces exactly $\sec x\tan x$ , the antiderivative of $\sec x\tan x$ is $\sec x$ .

\int \sec x\tan x\,dx = \sec x + C

Check it by differentiating

The derivative of $\sec x = \frac{1}{\cos x}$ is $\frac{\sin x}{\cos^2 x} = \sec x\tan x$ , so differentiating the answer returns the integrand at once.

Where the integral of sec x tan x shows up on the AP exam

It belongs to the basic antiderivatives of Topic 6.8, and Unit 6 is 15 to 20 percent of the exam. The pattern matters most when it sits inside a larger integrand, because you have to spot that $\sec x\tan x$ is a single antiderivative rather than a product to be broken up.

\int_{0}^{\pi/3} \sec x\tan x\,dx = \left[\sec x\right]_{0}^{\pi/3} = 2 - 1 = 1

It is also half of the pair used to set up the harder integral $\int \sec x\,dx$ , where multiplying by $\frac{\sec x + \tan x}{\sec x + \tan x}$ makes $\sec x\tan x$ appear in the numerator.

Common mistakes with the integral of sec x tan x

Answering $\tan x$ . The antiderivative of $\sec^2 x$ is $\tan x$ ; the antiderivative of $\sec x\tan x$ is $\sec x$ .
Adding a minus sign. Unlike the cosecant and cotangent pair, this one has none: $\int \sec x\tan x\,dx = \sec x + C$ .
Splitting the product. There is no way to integrate $\sec x$ and $\tan x$ separately and multiply; the product as a whole is the derivative of $\sec x$ .
Writing $\frac{\sec^2 x}{2}$ by treating it like a power. $\sec x\tan x$ is not $u\,du$ for any obvious $u$ ; it is a recognized derivative.

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 x tan x?

It is $\sec x + C$ . The product $\sec x\tan x$ is the derivative of $\sec x$ , so reversing that derivative gives $\sec x$ .

Does the answer have a minus sign?

No. $\int \sec x\tan x\,dx = \sec x + C$ with no minus. The minus sign belongs to the cosecant pair, $\int \csc x\cot x\,dx = -\csc x + C$ .

How is it different from the integral of sec^2 x?

$\int \sec^2 x\,dx = \tan x + C$ , while $\int \sec x\tan x\,dx = \sec x + C$ . Different integrands come from different standard derivatives.