AP Calculus AB and BC

Integral of cos x / sin^2 x: The Answer Is -csc x

Q: What is the integral of cos x / sin^2 x?

Both -\csc x + C and -\frac{1}{\sin x} + C name the same antiderivative, and either one is a complete answer.

Q: Is -1/sin x an acceptable final answer?

Yes. It is the same function as -\csc x, so a grader accepts either. Converting to \csc notation is a matter of convention.

Q: Why is the answer negative?

Because the power rule on u^{-2} raises the exponent to -1 and divides by -1. The minus comes from the exponent, not from the substitution.

The integral of cos x over sin squared x is negative cosecant x, plus C. Substituting u equals sin x turns the integrand into u to the power negative two, and the resulting negative 1 over sin x is normally written as negative csc x.

\int \frac{\cos x}{\sin^{2}x}\,dx = -\csc x + C

One substitution, then the power rule

The numerator $\cos x$ is exactly what $\sin x$ differentiates to, and the denominator is nothing but a power of $\sin x$ . That pairing is the signal to set $u = \sin x$ , giving $du = \cos x\,dx$ and consuming the numerator whole.

\int \frac{\cos x}{\sin^{2}x}\,dx = \int u^{-2}\,du = \frac{u^{-1}}{-1} + C = -\frac{1}{\sin x} + C

That is already correct. Since $\frac{1}{\sin x} = \csc x$ by definition, the tidy form is $-\csc x + C$ .

Recognising a reciprocal trig function

The step people stall on is the last one. A fraction with $1$ on top looks unfinished, so students keep manipulating $-\frac{1}{\sin x}$ looking for something better. There is nothing better: it is a named function wearing its definition instead of its name.

Differentiating backwards confirms it. $\frac{d}{dx}\left(-\csc x\right) = \csc x\cot x = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = \frac{\cos x}{\sin^{2}x}$ .

The same integral in disguise

Split the integrand as (1/sin x)(cos x/sin x) and it reads csc x cot x, a standard derivative you already know in reverse. Spotting that form gives the answer with no substitution at all.

The mistakes students make

Two of these are sign or pattern slips; the third is a reflex triggered by seeing a fraction.

Answering $\csc x$ with no minus sign. The power rule sends $u^{-2}$ to $\frac{u^{-1}}{-1}$ , and that negative belongs in the final answer.
Confusing the integrand with $\csc^{2}x$ and answering $-\cot x$ . The numerator here is $\cos x$ , not $1$ , so the two problems are different.
Writing $\ln\left|\sin^{2}x\right|$ because the integrand is a quotient. A log needs the numerator to be the derivative of the denominator, and $\sin^{2}x$ differentiates to $2\sin x\cos x$ , not $\cos x$ .

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 cos x / sin^2 x?

Both $-\csc x + C$ and $-\frac{1}{\sin x} + C$ name the same antiderivative, and either one is a complete answer.

Is -1/sin x an acceptable final answer?

Yes. It is the same function as $-\csc x$ , so a grader accepts either. Converting to $\csc$ notation is a matter of convention.

Why is the answer negative?

Because the power rule on $u^{-2}$ raises the exponent to $-1$ and divides by $-1$ . The minus comes from the exponent, not from the substitution.