Introduction
Integration often reverses differentiation. For cosine, the key fact is the derivative rule for sine: when angles are measured in radians, the derivative of \(\sin x\) is \(\cos x\). Therefore sine provides a direct antiderivative of cosine.
The same idea works when cosine is multiplied by a constant. The coefficient remains in the antiderivative, and an arbitrary constant is added because differentiating any constant gives zero. These statements can be checked directly by differentiation.
Antiderivatives and Indefinite Integral Notation
For example, to say that \(\sin x\) is an antiderivative of \(\cos x\) means precisely that differentiating \(\sin x\) gives \(\cos x\). The derivative rule for sine states that this holds for every real \(x\), with the trigonometric functions understood to use radians.
It follows immediately that \(\sin x+C\) is also an antiderivative of \(\cos x\) for any constant \(C\), because the derivative of \(C\) is zero. In fact, on an interval, these are all the antiderivatives of cosine.
Why the Constant Multiple Rule Works
Suppose \(a\) is a constant. Differentiation takes a constant factor outside the derivative, so the derivative of \(a\sin x\) is \(a\cos x\). Adding any constant does not change that derivative. This gives an antiderivative for every constant multiple of cosine.
This statement includes \(a=0\). In that case, the function being integrated is zero, and \(0\sin x+C=C\) is constant, as required.
Why All Antiderivatives Differ by a Constant
The derivative calculation verifies that \(a\sin x+C\) is an antiderivative. To see why there are no other antiderivatives on an interval, compare any two antiderivatives of the same function.
Let \(F\) and \(G\) both be antiderivatives of \(a\cos x\) on an interval \(I\). Then \(F'(x)=a\cos x\) and \(G'(x)=a\cos x\) for every \(x\) in \(I\).
Set \(H=F-G\). Then \(H'(x)=F'(x)-G'(x)=a\cos x-a\cos x=0\) throughout \(I\).
For any two points \(u<v\) in \(I\), that theorem gives a point \(c\) between them such that \(H(v)-H(u)=H'(c)(v-u)=0\). Therefore \(H(v)=H(u)\).
Since any two points in the interval have the same value of \(H\), the function \(H\) is constant on the interval. Thus \(F-G=C\) for some constant \(C\). If one antiderivative is \(a\sin x\), every other one must have the form \(a\sin x+C\). This proves the stated family of antiderivatives.
Worked Examples
Worked Example: Find an Antiderivative of Cosine
Find the general antiderivative of \(\cos x\), then verify it.
The derivative of \(\sin x\) is \(\cos x\), so include an arbitrary constant \(C\):
Differentiate the proposed answer term by term:
The derivative matches the integrand, so \(\sin x+C\) is the general antiderivative.
Worked Example: Integrate a Negative Constant Multiple
Find the general antiderivative of \(-4\cos x\), and check the result.
Keep the constant multiplier \(-4\) in the antiderivative of cosine:
Differentiate to verify, showing how the sign and coefficient carry through:
This is exactly the given function, so the answer is correct. The negative coefficient does not require an extra sign change: it remains the coefficient of \(\sin x\).
Worked Example: Use a Value to Select One Antiderivative
Find the antiderivative \(F\) of \(\frac{3}{2}\cos x\) that satisfies \(F(\frac{\pi}{6})=1\).
First write the family of antiderivatives:
At \(x=\frac{\pi}{6}\), the sine value is \(\sin(\frac{\pi}{6})=\frac{1}{2}\). Substitute the given function value and solve for \(C\):
Thus the particular antiderivative is \(F(x)=\frac{3}{2}\sin x+\frac{1}{4}\). Check both requirements. Its derivative is
and its value at the specified input is
Both the derivative condition and the prescribed value are satisfied.
A Common Sign Mistake
A frequent error is to attach a minus sign to the antiderivative of cosine, perhaps by confusing it with a different trigonometric derivative. The incorrect calculation is
Check it by differentiation rather than relying on memory:
which is not \(\cos x\). The proposed function is therefore an antiderivative of \(-\cos x\), not of \(\cos x\). The correction is to use the derivative rule \(\frac{d}{dx}\sin x=\cos x\): the antiderivative is \(\sin x+C\).
A second small but important issue is omitting \(C\) when the question asks for the general antiderivative. The expression \(\sin x\) is one antiderivative, but it does not describe the entire family. Since \(\frac{d}{dx}(\sin x+7)=\cos x\), for instance, other constants give other valid antiderivatives.
A Reliable Verification Routine
For an antiderivative involving a constant multiple of cosine, a short check can catch errors before they become embedded in a longer calculation.
- Identify the coefficient multiplying \(\cos x\), including its sign.
- Replace \(\cos x\) with \(\sin x\), retaining that same coefficient.
- Add \(C\) when writing the general antiderivative.
- Differentiate the proposed answer and compare the result with the original integrand.
This routine depends on the derivative rule for sine and the rule for differentiating a constant multiple. The final comparison is decisive: if the derivative does not reproduce the integrand exactly, the proposed antiderivative is not correct.
Why This Result Matters
Cosine is one of the basic functions whose antiderivative can be recognized directly from a derivative rule. That makes it a useful building block: expressions that are constant multiples of cosine can be integrated without changing the form of the trigonometric function, apart from replacing cosine by sine.
The central habit is to connect integration to differentiation. Rather than memorizing a sign in isolation, test the proposed answer by differentiating it. For cosine, this check shows why the antiderivative has a positive sine, why a coefficient stays in place, and why an arbitrary constant may be added.
Check Your Understanding
For each question, use differentiation to check any proposed antiderivative.
- What is the general antiderivative of \(\cos x\)?
- Find and verify the general antiderivative of \(6\cos x\).
- Differentiate \(-2\sin x+C\). Which function is it an antiderivative of?
- Why does the general antiderivative include an arbitrary constant?
- Find the antiderivative of \(-\frac{1}{3}\cos x\) that has value \(0\) at \(x=0\), and verify both conditions.
