Why the Signs Need Attention
The basic antiderivatives of sine and cosine look similar, but their signs are different. That difference is easy to overlook: the antiderivative of sine has a minus sign, whereas the antiderivative of cosine does not.
The signs are not rules to memorize without explanation. They follow from the derivative rules for sine and cosine. To find an antiderivative, reverse a derivative rule; to check the result, differentiate it and see whether the original function returns.
Antiderivatives and the Derivative Rules
A function \(F\) is an antiderivative of a function \(f\) on an interval if \(F'(x)=f(x)\) at every point in that interval. Thus, finding an antiderivative means asking which function differentiates to the integrand.
The derivative rules used here are
The first rule immediately identifies sine as an antiderivative of cosine. For sine, the second rule contains an extra negative sign. Multiplying cosine by \(-1\) reverses that sign, because the derivative of a constant multiple is the same constant multiple of the derivative.
An indefinite integral denotes the family of antiderivatives, not just one function. If \(F\) is an antiderivative, then \(F+C\) is also one for any constant \(C\), since the derivative of a constant is zero.
Deriving the Two Sign Patterns
For sine, start with the derivative of cosine. It gives \(-\sin x\), so multiplying cosine by \(-1\) gives a function whose derivative is \(\sin x\):
For cosine, the derivative of sine gives the required function directly:
These calculations establish one antiderivative for each function. They also explain why the signs differ: the derivative rule for cosine already includes a negative sign, while the derivative rule for sine does not.
To see why the constant accounts for all possibilities, suppose \(F\) and \(G\) are both antiderivatives of the same function on an interval. Then
A differentiable function with derivative zero throughout an interval is constant there. Indeed, for any two points \(a\) and \(b\) in the interval, the Mean Value Theorem gives a point \(c\) between them such that
So \(F-G\) has the same value at any two points in the interval. Therefore, two antiderivatives of the same function differ only by a constant. This is why each indefinite integral includes \(+C\).
The Basic Antiderivatives
Combining the derivative checks with the arbitrary-constant rule gives the two formulas:
These formulas apply to the functions \(\sin x\) and \(\cos x\) with argument \(x\). Each can be checked independently by differentiating its right-hand side. The constant disappears, and the original integrand remains.
| Integrand | Antiderivative family | Derivative check |
|---|---|---|
| \(\sin x\) | \(-\cos x+C\) | \(\frac{d}{dx}(-\cos x+C)=\sin x\) |
| \(\cos x\) | \(\sin x+C\) | \(\frac{d}{dx}(\sin x+C)=\cos x\) |
Worked Examples
Worked Example: Find an Antiderivative of Sine
Find \(\int\sin x\,dx\), and verify the result.
The derivative rule for cosine is \(\frac{d}{dx}\cos x=-\sin x\). To obtain positive sine as the derivative, place a negative sign in front of cosine:
Check by differentiating the proposed antiderivative:
The derivative is the original integrand, so the antiderivative and its sign are correct.
Worked Example: Find an Antiderivative of Cosine
Find \(\int\cos x\,dx\), and verify the result.
Since the derivative of sine is cosine, the antiderivative is sine, with an arbitrary constant added:
Differentiate to check:
The result differentiates to the integrand, so no extra negative sign belongs in this formula.
Worked Example: Use an Initial Value to Determine the Constant
Find the antiderivative \(F\) of \(\sin x\) that satisfies \(F(0)=3\).
Every antiderivative of \(\sin x\) has the form
Substitute \(x=0\) and use \(\cos 0=1\):
Add \(1\) to both sides to obtain \(C=4\). Therefore,
Check both requirements. Differentiating gives
and substituting \(x=0\) gives
Thus the derivative is \(\sin x\), and the specified value is satisfied.
A Reliable Sign Check
When a sign is uncertain, do not guess from memory. Differentiate the proposed antiderivative and compare the result with the integrand. This check uses precisely the derivative rules that produced the antiderivative formulas.
Decide whether the function to be integrated is \(\sin x\) or \(\cos x\).
Cosine differentiates to negative sine; sine differentiates to cosine.
Add \(C\), since differentiating any constant gives zero.
The result must equal the original integrand, including its sign.
Common Mistake: Losing the Negative Sign
A frequent error is to write \(\int\sin x\,dx=\cos x+C\). This mistakes the derivative of cosine for positive sine. In fact, differentiating that proposed answer gives
which is the negative of the required integrand. The correction is to put a negative sign before cosine:
A useful distinction is that the sign appears in the derivative rule for cosine, not in the derivative rule for sine. Therefore, it is the antiderivative of sine that carries the negative sign.
Why This Pattern Matters
The two formulas are a simple example of a broader habit in integration: an antiderivative should be understood by checking what its derivative does. For these basic trigonometric functions, that habit prevents a sign error before it becomes part of a longer calculation.
Keep the pair together, and remember the reason for the difference:
The first arrow needs a minus sign because \((\cos x)'=-\sin x\). The second does not, because \((\sin x)'=\cos x\). Differentiating each result confirms the pattern.
Check Your Understanding
For each question, use the derivative rules to justify the sign, not just recall the formula.
- What is the general form of an antiderivative of \(\sin x\)?
- What is the general form of an antiderivative of \(\cos x\)?
- Differentiate \(-\cos x+C\). Which sign does the result have?
- Why does \(\cos x+C\) fail to be an antiderivative of \(\sin x\)?
- Find the antiderivative \(F\) of \(\cos x\) satisfying \(F(0)=2\), and check both conditions.
