Tutorials › Calculus II › Checking Sine-Cosine Antiderivatives by Differentiation

Sine and Cosine Powers · Tutorial 125 of 998

Checking Sine-Cosine Antiderivatives by Differentiation

Use differentiation to test proposed antiderivatives of sine-cosine power products and pinpoint errors in their coefficients, exponents, or angle factors.

Beginner 9 min read

What You'll Learn

  • Apply the derivative test to a proposed sine-cosine antiderivative.
  • Differentiate powers of sine and cosine using the product and chain rules.
  • Check coefficients and exponents by comparing the derivative with the integrand.
  • Track the extra factor produced by a multiple-angle argument.
  • Correct a proposed answer that fails the derivative check.

Why Check by Differentiation?

An antiderivative can look plausible and still contain a small error. A missing coefficient, an incorrect exponent, or a forgotten factor from a multiple-angle argument can change the derivative—and therefore change the function being integrated.

Differentiation gives a direct test. If a proposed function differentiates to the stated integrand throughout the interval under consideration, it is an antiderivative there. For products of sine and cosine powers, the product rule and chain rule make this test systematic.

Definition: A function \(F\) is an antiderivative of \(f\) on an interval \(I\) if \(F\) is differentiable on \(I\) and \(F'(x)=f(x)\) for every \(x\in I\).

The task here is not to develop a method for finding every such antiderivative. It is to check a proposed answer carefully: differentiate it, simplify the result, and compare it with the integrand.

The Derivative Pattern for Sine and Cosine Powers

Let \(\theta=kx+\phi\), where \(k\) and \(\phi\) are constants. For nonnegative integer powers \(m\) and \(n\), the product and chain rules give the following derivative when \(m,n>0\):

$$ \frac{d}{dx}\left(\sin^m\theta\cos^n\theta\right) = k\left(m\sin^{m-1}\theta\cos^{n+1}\theta - n\sin^{m+1}\theta\cos^{n-1}\theta\right). $$

To see the two contributions, differentiate the sine power while leaving the cosine power unchanged, then differentiate the cosine power while leaving the sine power unchanged:

$$ \begin{aligned} \frac{d}{dx}\left(\sin^m\theta\cos^n\theta\right) &= m\sin^{m-1}\theta\,(k\cos\theta)\cos^n\theta + \sin^m\theta\,n\cos^{n-1}\theta\,(-k\sin\theta)\\ &= km\sin^{m-1}\theta\cos^{n+1}\theta - kn\sin^{m+1}\theta\cos^{n-1}\theta. \end{aligned} $$

The factor \(k\) comes from \(\frac{d\theta}{dx}=k\). The minus sign in the second term comes from the derivative of cosine. If one exponent is zero, differentiate the nonconstant factor directly; the displayed formula can also be interpreted by omitting the term multiplied by that zero exponent.

A proposed answer may be a sum of several terms, or it may involve a multiple-angle function. Differentiate every term, preserve all coefficients, and simplify only after the differentiation is complete. Then compare the simplified result with the original integrand.

Worked Example: Checking a Polynomial in Cosine

Worked Example: Checking a Polynomial in Cosine

Check the proposed antiderivative \(F(x)=-\frac{\cos^3 x}{3}+\frac{\cos^5 x}{5}\) for the integrand \(\sin^3 x\cos^2 x\).

Differentiate each term. By the chain rule, \(\frac{d}{dx}(\cos^3 x)=3\cos^2 x(-\sin x)\), so

$$ \frac{d}{dx}\left(-\frac{\cos^3 x}{3}\right) = -\frac{1}{3}\left(3\cos^2 x(-\sin x)\right) = \sin x\cos^2 x. $$

For the other term, \(\frac{d}{dx}(\cos^5 x)=5\cos^4 x(-\sin x)\), and therefore

$$ \frac{d}{dx}\left(\frac{\cos^5 x}{5}\right) = \frac{1}{5}\left(5\cos^4 x(-\sin x)\right) = -\sin x\cos^4 x. $$

Adding the derivatives and factoring gives

$$ \begin{aligned} F'(x) &=\sin x\cos^2 x-\sin x\cos^4 x\\ &=\sin x\cos^2 x(1-\cos^2 x)\\ &=\sin x\cos^2 x\sin^2 x\\ &=\sin^3 x\cos^2 x. \end{aligned} $$

The identity \(1-\cos^2 x=\sin^2 x\) shows that the derivative equals the stated integrand. Thus the proposed function passes the check.

Follow the Angle Through the Chain Rule

When a sine or cosine contains \(kx+\phi\), differentiating the inside contributes a factor of \(k\). That factor must be matched when checking the coefficient in front of the proposed antiderivative. The next example isolates this issue.

Worked Example: Checking a Multiple-Angle Power

Check whether \(F(x)=\frac{\sin^3(2x)}{6}\) is an antiderivative of \(\sin^2(2x)\cos(2x)\).

Set \(\theta=2x\). Then \(\frac{d\theta}{dx}=2\). Applying the chain rule to the cube gives

$$ \begin{aligned} F'(x) &=\frac{1}{6}\cdot 3\sin^2(2x)\cdot\frac{d}{dx}\bigl(\sin(2x)\bigr)\\ &=\frac{1}{6}\cdot 3\sin^2(2x)\cdot 2\cos(2x)\\ &=\sin^2(2x)\cos(2x). \end{aligned} $$

The derivative matches the integrand exactly, so \(F\) is an antiderivative. The coefficient \(1/6\) is verified by the three factors \(1/6\), \(3\), and \(2\) in the derivative.

Check Multiple-Angle Terms Carefully

Multiple-angle identities can produce proposed antiderivatives containing terms such as \(\sin(4x)\). Their coefficients must account for both the coefficient of the sine term and the derivative of its angle. Differentiate the proposed expression directly rather than guessing whether a factor has been included.

Worked Example: Checking a Fourfold-Angle Term

Check the proposed antiderivative \(F(x)=\frac{x}{8}-\frac{\sin(4x)}{32}\) for \(\sin^2 x\cos^2 x\).

The derivative of the first term is \(1/8\). For the second, the chain rule gives \(\frac{d}{dx}\sin(4x)=4\cos(4x)\). Hence

$$ \begin{aligned} F'(x) &=\frac{1}{8}-\frac{1}{32}\left(4\cos(4x)\right)\\ &=\frac{1}{8}-\frac{\cos(4x)}{8}\\ &=\frac{1-\cos(4x)}{8}. \end{aligned} $$

The double-angle identity \(\cos(2x)=1-2\sin^2 x\), applied to \(2x\), gives \(\cos(4x)=1-2\sin^2(2x)\). Also, \(\sin(2x)=2\sin x\cos x\). Therefore

$$ \begin{aligned} \frac{1-\cos(4x)}{8} &=\frac{2\sin^2(2x)}{8}\\ &=\frac{\sin^2(2x)}{4}\\ &=\frac{4\sin^2 x\cos^2 x}{4}\\ &=\sin^2 x\cos^2 x. \end{aligned} $$

Thus the derivative equals the integrand, and the proposed antiderivative is correct. In particular, differentiating \(\sin(4x)\) supplies a factor of \(4\), which is essential to the check.

A Reliable Checking Procedure

A derivative check is most useful when it is done in a fixed order. The following steps help separate the chain-rule calculation from the algebraic comparison.

1
Write down the proposed function.
Keep its coefficients and exponents visible; do not silently alter them to make the answer fit.
2
Differentiate term by term.
Use the product rule for products and the chain rule for each sine or cosine of a nontrivial angle.
3
Simplify the derivative.
Factor common powers where useful and apply an identity only when it helps compare the result with the integrand.
4
Compare with the integrand.
Check the entire expression, including its sign, numerical coefficient, exponents, and angle.
5
Correct any mismatch and check again.
A correction is complete only when differentiating the revised expression gives the integrand.

Why the Derivative Test Is Decisive

The definition of antiderivative makes the first part of the test immediate: if \(F'(x)=f(x)\) at every point of an interval, then \(F\) is an antiderivative of \(f\) on that interval. The test also explains why two correct answers can differ by a constant.

Suppose \(F\) and \(G\) are differentiable on an interval \(I\) and both have derivative \(f\). Define \(H=F-G\). Then, for every \(x\in I\),

$$ H'(x)=F'(x)-G'(x)=f(x)-f(x)=0. $$

For any two points \(a,b\in I\) with \(a<b\), the Mean Value Theorem applies to \(H\) on \([a,b]\): \(H\) is continuous on that closed interval and differentiable on its interior. It gives a point \(c\in(a,b)\) such that

$$ H(b)-H(a)=H'(c)(b-a)=0. $$

Thus \(H(b)=H(a)\). Since the points were arbitrary, \(H\) is constant on \(I\), so \(F\) and \(G\) differ by a constant there. This is why checking the derivative, rather than comparing two formulas term by term, is the decisive test.

Common Pitfall: Omitting the Multiple-Angle Factor

Consider \(\sin^2(3x)\cos(3x)\). A tempting proposed antiderivative is \(\frac{\sin^3(3x)}{3}\), but this omits the effect of differentiating \(3x\). Differentiate the proposal to see the exact error:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{\sin^3(3x)}{3}\right) &=\frac{1}{3}\cdot 3\sin^2(3x)\cdot 3\cos(3x)\\ &=3\sin^2(3x)\cos(3x). \end{aligned} $$

The derivative is three times the desired integrand, not the integrand itself. To correct the coefficient, use \(\frac{\sin^3(3x)}{9}\). Checking this revision:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{\sin^3(3x)}{9}\right) &=\frac{1}{9}\cdot 3\sin^2(3x)\cdot 3\cos(3x)\\ &=\sin^2(3x)\cos(3x). \end{aligned} $$

The revised expression passes the derivative test. The correction comes from accounting for both the derivative of the cube and the factor \(3\) from the inner angle.

What the Check Can Tell You

If the derivative has the right powers and angles but the wrong numerical multiple, inspect the proposed coefficient and any chain-rule factor. If a power is wrong, check how differentiating a power changes its exponent: differentiating \(\sin^m x\) lowers the sine exponent by one and introduces a cosine, while differentiating \(\cos^n x\) lowers the cosine exponent by one, introduces a sine, and contributes a minus sign.

Sometimes the differentiated expression does not look like the integrand until it is factored or simplified with a trigonometric identity. That is not a reason to declare the candidate incorrect. Compare equivalent expressions, but do not use an identity to conceal a missing factor or sign. A successful check ends with an explicit equality between the derivative and the stated integrand.

Key takeaway: Differentiate the proposed expression exactly as written. Track signs, coefficients, exponents, and inner-angle factors before simplifying, then verify that the result equals the integrand.

Check Your Understanding

For each question, use differentiation to justify your answer rather than relying on the appearance of the proposed expression.

  1. What condition must a function \(F\) satisfy to be an antiderivative of \(f\) on an interval?
  2. Differentiate \(\sin^4(2x)\). Which factor comes from differentiating the inner angle?
  3. Does \(\frac{\sin^3(5x)}{15}\) differentiate to \(\sin^2(5x)\cos(5x)\)? Show the derivative and decide.
  4. Why can two antiderivatives of the same function on an interval differ only by a constant?
  5. When checking a multiple-angle expression, which factors and signs should be checked before simplifying?

Test yourself: Calculus II final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon