Standard Antiderivatives of Elementary Functions
Five integrals come up so often that you simply memorize them: \(\int \sin x\,dx = -\cos x + C\), \(\int \cos x\,dx = \sin x + C\), \(\int \sec^2 x\,dx = \tan x + C\), \(\int e^x\,dx = e^x + C\), and \(\int \frac{1}{x}\,dx = \ln|x| + C\). Each one is a derivative rule you already know, run backward — and each one can be checked by differentiating the answer.
By the end you'll be able to recall all five standard forms — with the right sign on \(\int \sin x\,dx\) and the absolute value on \(\int \frac{1}{x}\,dx\) — and verify any claimed antiderivative by differentiating it.
Predict: is the antiderivative of \(\cos x\) equal to \(\sin x\), or \(-\sin x\)? The checker starts on \(\int \cos x\,dx\) — drag the point along the plotted answer curve and watch the tangent slope readout trace out \(\cos x\) itself. Then switch to \(\int \sin x\,dx\) and see why that answer needs a minus sign.
Pick a table entry below. The solid curve is the answer F(x); the dashed curve is the integrand f(x) you started from. At the draggable point, the tangent to F has a slope — and that slope lands exactly on the dashed curve, every time: differentiating the answer gives back the function you integrated. The C slider shifts F up and down; notice the slope never changes — that's why every entry carries "+ C".
Answer plotted: F(x) = sin x + C. Tangent slope at x = 1.00: 0.540 · integrand cos x there: 0.540 — they match.
The standard antiderivative table is the derivative table read backward — five entries you'll use constantly, each certified by one differentiation.
The table
Here are the five forms, each with the derivative rule that certifies it. Every entry carries \(+\,C\), because adding a constant never changes a derivative:
| \(\int \sin x\,dx = -\cos x + C\) | because \(\frac{d}{dx}(-\cos x) = \sin x\) |
| \(\int \cos x\,dx = \sin x + C\) | because \(\frac{d}{dx}\sin x = \cos x\) |
| \(\int \sec^2 x\,dx = \tan x + C\) | because \(\frac{d}{dx}\tan x = \sec^2 x\) |
| \(\int e^x\,dx = e^x + C\) | because \(\frac{d}{dx}e^x = e^x\) |
| \(\int \frac{1}{x}\,dx = \ln|x| + C\) | because \(\frac{d}{dx}\ln|x| = \frac{1}{x}\) for \(x \neq 0\) |
Nothing here is new — it's your derivative table run in reverse. You already know the trig derivatives \(\frac{d}{dx}\sin x = \cos x\) and \(\frac{d}{dx}\cos x = -\sin x\); reversing the first gives \(\int \cos x\,dx = \sin x + C\) directly, but reversing the second takes one extra move: since \(\frac{d}{dx}\cos x\) produces \(-\sin x\), you must negate to land on \(+\sin x\), so \(\int \sin x\,dx = -\cos x + C\). That is where the most common sign error in calculus lives — writing \(+\cos x\) and failing the differentiation check, since \(\frac{d}{dx}\cos x = -\sin x \neq \sin x\). The exponential is the fixed point of differentiation (\(\frac{d}{dx}e^x = e^x\)), so it's its own antiderivative. And \(\int \frac{1}{x}\,dx = \ln|x| + C\) plugs the one hole the reverse power rule leaves: \( \int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) breaks down at \(n = -1\) (division by zero), and the logarithm steps in exactly there.
Evaluate \(\displaystyle\int_0^{\pi/2} \cos x\,dx\). Step 1: from the table, an antiderivative of \(\cos x\) is \(\sin x\). Step 2: apply the FTC: \([\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = \) 1. The area under the right half of one hump of the cosine — from 0 to \(\pi/2\) — is exactly 1.
Evaluate \(\displaystyle\int_1^{e^2} \frac{1}{x}\,dx\). From the table, an antiderivative of \(\frac{1}{x}\) is \(\ln|x|\), and both limits are positive so \(\ln|x| = \ln x\) here. Now finish it: \([\ln x]_1^{e^2} = \ln(e^2) - \ln(1) = \) ____
Reveal the answer
\(\ln(e^2) = 2\) and \(\ln(1) = 0\), so the integral is \(2 - 0 = \) 2. Quick check by differentiating: \(\frac{d}{dx}\ln x = \frac{1}{x}\) — the integrand — so the antiderivative was right before we ever touched the limits.
More info — why the absolute value in ∫ (1/x) dx really matters
On the positive axis, \(\frac{d}{dx}\ln x = \frac{1}{x}\) and all is well. But \(\frac{1}{x}\) also exists for negative x, where \(\ln x\) doesn't. On that side, \(|x| = -x\), so \(\ln|x| = \ln(-x)\), and the chain rule gives \(\frac{d}{dx}\ln(-x) = \frac{1}{-x}\cdot(-1) = \frac{1}{x}\). One formula, \(\ln|x| + C\), therefore works on both sides of zero — try it in the checker above by dragging the point to negative x on the \(\int 1/x\,dx\) entry. Note what it does not fix: \(x = 0\) stays excluded, since neither \(\frac{1}{x}\) nor \(\ln|x|\) is defined there. Paul's Online Notes (linked in Dive deeper below) works several more examples with this entry.
These five forms are the raw material for everything that follows: substitution turns integrals with shifted or scaled arguments — like \(\int \cos(3x)\,dx\) — back into table entries, and the table plus the FTC is how you'll evaluate the definite integrals in the applications ahead.
Check your understanding
What is \(\int \sec^2 x\,dx\)?
Evaluate \(\displaystyle\int_0^1 e^x\,dx\).
Why is \(\int \frac{1}{x}\,dx = \ln|x| + C\) rather than \(\ln x + C\)?
A classmate writes \(\int \sin x\,dx = \cos x + C\). Check the claim by differentiating. What do you find?
Recap
- \(\int \sin x\,dx = -\cos x + C\) — the minus sign comes from \(\frac{d}{dx}\cos x = -\sin x\); this is the classic sign trap.
- \(\int \cos x\,dx = \sin x + C\) and \(\int \sec^2 x\,dx = \tan x + C\).
- \(\int e^x\,dx = e^x + C\) — the exponential is its own antiderivative.
- \(\int \frac{1}{x}\,dx = \ln|x| + C\) — the absolute value makes the formula valid for negative x too, filling the reverse power rule's gap at \(n = -1\).
- Every entry is certified the same way: differentiate the answer and you must get the integrand back.
Dive deeper
- Paul's Online Notes — Computing Indefinite Integrals List the standard antiderivatives of trig, exponential, and reciprocal functions
Sources
- Standard Antiderivatives of Elementary Functions