Integral of sin x: Answer, Proof, and Mistakes

Why the antiderivative of sin x carries a minus sign

Antidifferentiation is differentiation run backwards, so the fastest way to find $\int \sin x\,dx$ is to ask which function has $\sin x$ as its derivative. The trig derivative that produces a sine is the one for cosine, and it comes with a negative sign.

That is off by a factor of $-1$, so multiply by $-1$. Differentiating $-\cos x$ cancels the two negatives and leaves exactly $\sin x$.

Because the check works for every real $x$, the antiderivative is valid on the whole real line. There is no domain restriction to state, unlike $\int \frac{1}{x}\,dx$.

Definite integrals of sin x you should recognize

Antiderivatives of $\sin x$ appear on the AP exam mostly inside definite integrals evaluated with the Fundamental Theorem of Calculus (Topic 6.7). Unit 6, Integration and Accumulation of Change, is weighted 15 to 20 percent on both AB and BC, so these evaluations are worth drilling until they are automatic.

One arch of the sine curve is the standard case. Substituting the limits into $-\cos x$ gives a clean whole number, which is why this integral shows up so often in area and average-value problems.

Over a full period the answer is zero, because the arch below the axis contributes the same amount with the opposite sign. This is the difference between net signed area and total area: for total area you would integrate $|\sin x|$, or double the result over $[0, \pi]$.

The same antiderivative feeds Unit 8. The average value of $\sin x$ on $[0, \pi]$ (Topic 8.1) is the integral divided by the length of the interval, and it is a number worth remembering.

• $\int_0^{\pi/2} \sin x\,dx = 1$, the quarter arch.
• $\int_0^{\pi} \sin x\,dx = 2$, one full arch.
• $\int_0^{2\pi} \sin x\,dx = 0$, one full period of net signed area.
• The average value of $\sin x$ on $[0, \pi]$ is $\frac{2}{\pi} \approx 0.6366$.

When the sine has something else inside it

A bare $\sin x$ is rare. Most AP integrals hand you $\sin(\text{something})$, and then the rule is substitution (Topic 6.9), not the memorized form. The tell is that differentiating $-\cos(3x)$ would produce an extra factor of 3 by the chain rule, so the antiderivative has to divide it back out.

The general linear case follows the same pattern for any nonzero constant $k$. This shortcut is safe only when the inside is linear; anything else needs a real substitution.

When the inside is not linear, look for its derivative sitting in the integrand. In $\int x\sin(x^2)\,dx$, let $u = x^2$, so $du = 2x\,dx$ and $x\,dx = \frac{1}{2}du$.

A squared sine is a different animal entirely. $\sin^2 x$ is not a composite of the form $\sin(u)$, so no substitution untangles it directly; you rewrite it with the power-reducing identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$ first.

Common mistakes with the integral of sin x

1. Dropping the minus sign. The derivative of $\sin x$ is $+\cos x$, but the integral of $\sin x$ is $-\cos x + C$. Going forwards and backwards move the sign in opposite directions, and mixing them up is the most common error on this integral.
2. Forgetting the constant of integration. An indefinite integral without $+ C$ is an incomplete answer.
3. Skipping the reciprocal factor on a composite. $\int \sin(3x)\,dx = -\frac{1}{3}\cos(3x) + C$, not $-\cos(3x) + C$. Differentiate your answer to catch this in seconds.
4. Sign-flipping the limits. In $\left[-\cos x\right]_0^{\pi}$ you compute $(-\cos\pi) - (-\cos 0)$, which is $1 + 1 = 2$. Distributing the outer minus sign carelessly turns a 2 into a 0.
5. Reporting 0 as the area under one full period. $\int_0^{2\pi} \sin x\,dx = 0$ is the net signed area; the total area is 4.
6. Working in degrees. Every antiderivative here assumes radians, so a calculator left in degree mode will disagree with your exact answer.