AP Calculus AB and BC

Integral of csc x cot x: Answer and Proof

Q: What is the integral of csc x cot x?

It is -\csc x + C. Since \frac{d}{dx}\csc x = -\csc x\cot x, differentiating -\csc x gives \csc x\cot x, so that is the antiderivative.

Q: Why is there a minus sign?

Because the cosecant derivative already has one: \frac{d}{dx}\csc x = -\csc x\cot x. Reversing a derivative that produced a negative means negating, so -\csc x is the antiderivative of +\csc x\cot x.

Q: How does it compare to the integral of sec x tan x?

They are mirror images. \int \sec x\tan x\,dx = \sec x + C has no minus sign, while \int \csc x\cot x\,dx = -\csc x + C does, because the cosecant derivative carries the minus.

The integral of csc x cot x with respect to x is -csc x + C. The product csc x cot x is the negative of the derivative of csc x, since d/dx csc x = -csc x cot x, so the antiderivative is -csc x. Differentiating -csc x returns csc x cot x.

\int \csc x\cot x\,dx = -\csc x + C

How to integrate csc x cot x

Like the other reciprocal-trig products, this one is read straight off the derivative table. You want the function whose derivative is $\csc x\cot x$ , and the cosecant rule supplies it.

\frac{d}{dx}\csc x = -\csc x\cot x

The derivative of $\csc x$ is $-\csc x\cot x$ , so to land on a positive $\csc x\cot x$ you differentiate $-\csc x$ instead. That makes $-\csc x$ the antiderivative.

\int \csc x\cot x\,dx = -\csc x + C

Check it by differentiating

Differentiate $-\csc x$ : the derivative of $\csc x$ is $-\csc x\cot x$ , so the derivative of $-\csc x$ is $\csc x\cot x$ , the integrand. The minus in the answer cancels the minus in the derivative rule.

Where the integral of csc x cot x shows up on the AP exam

It is a Topic 6.8 basic antiderivative, part of the 15 to 20 percent that Unit 6 carries. It is the mirror image of $\int \sec x\tan x\,dx = \sec x + C$ ; the only differences are the co-functions and the minus sign.

\int_{\pi/6}^{\pi/2} \csc x\cot x\,dx = \left[-\csc x\right]_{\pi/6}^{\pi/2} = -1 - (-2) = 1

As with $\csc^2 x$ , the domain matters. $\csc x\cot x$ is undefined where $\sin x = 0$ , so a definite integral straddling a multiple of $\pi$ is improper.

Common mistakes with the integral of csc x cot x

Dropping the minus sign and writing $\csc x + C$ . Differentiating $\csc x$ gives $-\csc x\cot x$ , the wrong sign.
Confusing it with $\int \sec x\tan x\,dx = \sec x + C$ . The secant pair has no minus; the cosecant pair does.
Answering $-\cot x$ . That is the antiderivative of $\csc^2 x$ , not of $\csc x\cot x$ .
Integrating across $\sin x = 0$ . On any interval containing a multiple of $\pi$ the integrand has a vertical asymptote and the integral diverges.

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 csc x cot x?

It is $-\csc x + C$ . Since $\frac{d}{dx}\csc x = -\csc x\cot x$ , differentiating $-\csc x$ gives $\csc x\cot x$ , so that is the antiderivative.

Why is there a minus sign?

Because the cosecant derivative already has one: $\frac{d}{dx}\csc x = -\csc x\cot x$ . Reversing a derivative that produced a negative means negating, so $-\csc x$ is the antiderivative of $+\csc x\cot x$ .

How does it compare to the integral of sec x tan x?

They are mirror images. $\int \sec x\tan x\,dx = \sec x + C$ has no minus sign, while $\int \csc x\cot x\,dx = -\csc x + C$ does, because the cosecant derivative carries the minus.