AP Calculus AB and BC

Integral of sin x / cos^3 x: Two Minus Signs Cancel

Q: What is the integral of sin x / cos^3 x?

The antiderivative is \frac{1}{2\cos^{2}x} + C, which is normally written \frac{\sec^{2}x}{2} + C.

Q: I got tan^2(x)/2 instead. Is that wrong?

No. Rewriting the integrand as \tan x\sec^{2}x and substituting u = \tan x gives that form, and since \sec^{2}x = \tan^{2}x + 1 the two answers differ by \frac{1}{2}. The constant of integration absorbs the difference, so both are correct.

Q: Why does the minus sign disappear?

Because there are two of them. du = -\sin x\,dx contributes one and the power rule on u^{-3} contributes another, and a product of two negatives is positive.

The integral of sin x over cos cubed x is secant squared x over 2, plus C. Substituting u equals cos x brings a minus sign from du, and the power rule on u to the negative three brings a second one, so the two cancel and the answer comes out positive.

\int \frac{\sin x}{\cos^{3}x}\,dx = \frac{\sec^{2}x}{2} + C

Substitute, and track both minus signs

Take $u = \cos x$ . Then $du = -\sin x\,dx$ , so $\sin x\,dx = -du$ and the whole integral picks up a minus sign before the power rule is applied at all.

\int \frac{\sin x}{\cos^{3}x}\,dx = -\int u^{-3}\,du = -\frac{u^{-2}}{-2} + C = \frac{1}{2u^{2}} + C

Putting $u = \cos x$ back gives $\frac{1}{2\cos^{2}x}$ , and since $\frac{1}{\cos x} = \sec x$ , the finished form is $\frac{\sec^{2}x}{2} + C$ .

Why the answer is positive

Two negatives appear and they cancel: one from $du = -\sin x\,dx$ , one from the power rule dividing by $-2$ . Handle only one of them and the sign of the whole answer flips.

A graph check settles it in seconds. On $0 < x < \frac{\pi}{2}$ the integrand $\frac{\sin x}{\cos^{3}x}$ is positive, so any antiderivative must be increasing there. $\frac{\sec^{2}x}{2}$ climbs on that interval; $-\frac{\sec^{2}x}{2}$ falls, so the negative version is out.

A sign check worth building into your habits

Pick an interval where the integrand is clearly positive, here 0 to pi over 2, and ask whether your antiderivative is increasing on it. A flipped global sign shows up immediately, and this costs less time than redoing the substitution.

The mistakes students make

The first is the sign error the problem is built to catch. The other two come from mishandling the negative exponent or reaching for a log.

Answering $-\frac{\sec^{2}x}{2}$ , which is what you get by applying the minus from $du$ and forgetting that the power rule supplies a second one.
Moving the exponent the wrong way and writing $\int u^{-3}du = \frac{u^{-4}}{-4}$ , which leads to $\frac{1}{4\cos^{4}x}$ . The exponent goes up to $-2$ , not down to $-4$ .
Answering $\ln\left|\cos^{3}x\right|$ on the grounds that the integrand is a fraction. The derivative of $\cos^{3}x$ is $-3\cos^{2}x\sin x$ , nowhere near the $\sin x$ on top.

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 sin x / cos^3 x?

The antiderivative is $\frac{1}{2\cos^{2}x} + C$ , which is normally written $\frac{\sec^{2}x}{2} + C$ .

I got tan^2(x)/2 instead. Is that wrong?

No. Rewriting the integrand as $\tan x\sec^{2}x$ and substituting $u = \tan x$ gives that form, and since $\sec^{2}x = \tan^{2}x + 1$ the two answers differ by $\frac{1}{2}$ . The constant of integration absorbs the difference, so both are correct.

Why does the minus sign disappear?

Because there are two of them. $du = -\sin x\,dx$ contributes one and the power rule on $u^{-3}$ contributes another, and a product of two negatives is positive.