Introduction
An antiderivative reverses differentiation: if \(F'(x)=f(x)\), then \(F\) is an antiderivative of \(f\). For sine, the key fact is that the derivative of cosine is negative sine. That negative sign determines the answer.
In this tutorial, we use that derivative rule to find antiderivatives of \(\sin x\) and of constant multiples of \(\sin x\). The quickest way to check each answer is to differentiate it and compare the result with the function we started with.
Definition and the Basic Formula
We use radians for angles, as is standard in calculus. The derivative rule needed here is \(\frac{d}{dx}\cos x=-\sin x\). Therefore, applying the constant multiple rule for differentiation gives
Thus \(-\cos x\) is an antiderivative of \(\sin x\). Also, adding a constant does not change a derivative, because the derivative of a constant is zero. The whole family of antiderivatives is therefore
The indefinite integral notation on the left represents a family of antiderivatives, not a single function. Different choices of \(C\) give functions that differ only by a constant, and each has derivative \(\sin x\).
Why the Negative Sign Is Required
The sign is not a convention to memorize without explanation. It follows directly from the derivative of cosine. Differentiating \(\cos x\) produces \(-\sin x\), so \(\cos x\) itself has the wrong derivative for an antiderivative of positive sine. Negating cosine reverses that sign.
This check is also a useful way to remember the formula: differentiate the proposed answer and see whether it returns the original integrand. The required negative sign is visible in that verification.
Worked Examples
Worked Example: Find an Antiderivative of \(\sin x\)
Find the general antiderivative of \(f(x)=\sin x\).
From the derivative rule for cosine, \(\frac{d}{dx}(\cos x)=-\sin x\). Hence
All antiderivatives differ by a constant, so the general answer is
Check: Differentiating the answer gives \(F'(x)=\sin x\), as required.
Worked Example: Integrate a Negative Constant Multiple
Find the general antiderivative of \(f(x)=-4\sin x\).
By the constant multiple rule, a constant factor can be taken outside the integral. Applying the sine formula gives
Check: Differentiate the proposed answer term by term:
This matches the original function. Notice that the negative coefficient in the integrand combines with the negative sign in the sine antiderivative, producing positive \(4\cos x\).
Worked Example: Use a Value to Determine the Constant
Suppose \(F'(x)=\frac{3}{2}\sin x\) and \(F\left(\frac{\pi}{3}\right)=2\). Find \(F(x)\).
First find the general antiderivative:
Substitute \(x=\frac{\pi}{3}\) and \(F\left(\frac{\pi}{3}\right)=2\). Since \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\),
Adding \(\frac{3}{4}\) to both sides gives \(C=\frac{11}{4}\). Thus
Check the derivative:
Check the given value:
Both the derivative condition and the specified function value are satisfied.
Constant Multiples of Sine
Let \(k\) be any real constant. The constant multiple rule for integration and the basic sine formula give
The formula works whether \(k\) is positive, negative, or zero. When \(k=0\), it gives \(C\), which is correct because every constant function has derivative zero. When \(k\) is negative, the two negative signs—the one in the coefficient and the one in the basic antiderivative—must both be accounted for.
To verify the rule, differentiate the proposed expression:
Since its derivative is the given integrand, \(-k\cos x+C\) is an antiderivative. The constant \(C\) remains arbitrary because differentiating it contributes zero.
Why These Are All the Antiderivatives
Checking that a formula differentiates to the integrand proves it is an antiderivative. To see why the formula gives every antiderivative on an interval, let \(F\) be any antiderivative of \(\sin x\). Then \(F'(x)=\sin x\). Define \(G(x)=F(x)+\cos x\). By the sum rule and the derivative of cosine,
Now take any two points \(a,b\) in the interval. The mean value theorem applies to \(G\) between those points because \(G\) is continuous on the closed interval between them and differentiable in its interior. It gives a point \(c\) between \(a\) and \(b\) such that
Thus \(G(b)=G(a)\) for every pair of points in the interval, so \(G\) is constant there. Write that constant as \(C\). Since \(G(x)=F(x)+\cos x\), rearranging yields \(F(x)=-\cos x+C\). Therefore every antiderivative has the stated form, not merely the particular one we checked.
Common Sign Error
A frequent mistake is to omit the negative sign and write \(\int\sin x\,dx=\cos x+C\). This fails the defining test for an antiderivative:
which is the negative of the requested integrand. The correction is to negate cosine:
Another way this error appears is mishandling a negative coefficient. For example, integrating \(-4\sin x\) does not give \(-4\cos x+C\), because
which has the wrong sign. The correct answer is \(4\cos x+C\), whose derivative is \(-4\sin x\).
Why the Formula Is Useful
Antiderivatives allow us to reconstruct functions from their rates of change. If a quantity changes at a rate proportional to \(\sin x\), the formula identifies a function whose derivative is that rate. The arbitrary constant records the fact that a rate alone does not determine a unique function: adding a constant changes the function's level but leaves its derivative unchanged.
When an initial or boundary value is also known, that extra condition determines the constant, as in the third example. The main procedure remains compact: use the negative cosine formula, include the coefficient, retain \(C\), and verify by differentiation.
Check Your Understanding
For each question, use differentiation to check your reasoning.
- What is the general antiderivative of \(\sin x\), and why must its cosine term have a negative sign?
- Find the general antiderivative of \(6\sin x\).
- Find the general antiderivative of \(-\frac{2}{5}\sin x\).
- Does \(\cos x+C\) have derivative \(\sin x\)? Show the derivative that determines your answer.
- If \(F'(x)=2\sin x\) and \(F(0)=3\), find \(F(x)\) and check both conditions.
