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Sine and Cosine Powers · Tutorial 108 of 998

Integrating Even Powers of Sine

Use power reduction to turn even powers of sine into sums of cosine terms, then integrate and verify each term.

Beginner 9 min read

What You'll Learn

  • Apply the sine power-reduction identity to an even power.
  • Keep track of the constant term as well as the cosine terms.
  • Integrate cosine terms with their frequency factors included.
  • Evaluate integrals of second, fourth, and sixth powers of sine.
  • Check an antiderivative by differentiating it.

Why Even Powers Need a Different Approach

For an odd power of sine, one factor of \(\sin x\) can be saved while the remaining even power is rewritten using \(\sin^2 x=1-\cos^2 x\). For an even power, there is no single sine factor left over. A useful alternative is the power-reduction identity: it rewrites the square of sine using a cosine with twice the angle. Applying this identity lets us express an even power as a constant and cosine terms, each of which can be integrated directly.

The important bookkeeping is to preserve every coefficient and frequency, including the constant term. The constant integrates to a term involving \(x\); a cosine term integrates to a sine term whose coefficient must be divided by its frequency.

Key identity: The power-reduction identity for sine is \(\sin^2 x=\frac{1-\cos(2x)}{2}\). It follows from \(\cos(2x)=1-2\sin^2 x\).

From a Trigonometric Expansion to an Antiderivative

After power reduction, suppose the integrand has been written as a constant plus cosine terms. Each term has a direct antiderivative. In particular, the chain rule gives \(\frac{d}{dx}\sin(2kx)=2k\cos(2kx)\), where \(k\) is a positive integer. Therefore, when integrating a cosine term, its coefficient is divided by \(2k\).

$$ \int \cos(2kx)\,dx=\frac{\sin(2kx)}{2k}+C,\qquad k\ne 0. $$

More generally, if power reduction produces a finite expansion \(c_0+\sum_{k=1}^{m}a_k\cos(2kx)\), then integrating term by term gives the following antiderivative.

$$ c_0x+\sum_{k=1}^{m}\frac{a_k}{2k}\sin(2kx)+C. $$

This expression is useful for checking the bookkeeping: the constant contributes \(c_0x\), and each cosine term contributes a sine term with its own frequency divisor. The values of \(c_0\) and \(a_k\) come from the power-reduction expansion; they must not be guessed or discarded.

Worked Example: Integrating the Second Power

Worked Example: Integrating \(\sin^2 x\)

Apply the power-reduction identity directly:

$$ \sin^2 x=\frac{1-\cos(2x)}{2} =\frac12-\frac12\cos(2x). $$

Integrate the constant and cosine terms separately. Since \(\int \cos(2x)\,dx=\frac12\sin(2x)\), the coefficient \(-\frac12\) is also multiplied by \(\frac12\):

$$ \begin{aligned} \int \sin^2 x\,dx &=\int\left(\frac12-\frac12\cos(2x)\right)\,dx\\ &=\frac{x}{2}-\frac12\left(\frac{\sin(2x)}{2}\right)+C\\ &=\frac{x}{2}-\frac{\sin(2x)}{4}+C. \end{aligned} $$

Check by differentiation:

$$ \frac{d}{dx}\left(\frac{x}{2}-\frac{\sin(2x)}{4}\right) =\frac12-\frac{2\cos(2x)}{4} =\frac12-\frac12\cos(2x) =\sin^2 x. $$

The derivative returns the original integrand, so the antiderivative is correct.

Worked Example: Integrating the Fourth Power

Worked Example: Integrating \(\sin^4 x\)

Write \(\sin^4 x=(\sin^2 x)^2\), then apply the power-reduction identity to each squared sine factor:

$$ \begin{aligned} \sin^4 x &=\left(\frac{1-\cos(2x)}{2}\right)^2\\ &=\frac14\left(1-2\cos(2x)+\cos^2(2x)\right). \end{aligned} $$

Reduce the remaining squared cosine using \(\cos^2(2x)=\frac{1+\cos(4x)}{2}\). Substitute this into the expression and collect the constant and cosine terms:

$$ \begin{aligned} \sin^4 x &=\frac14\left(1-2\cos(2x)+\frac{1+\cos(4x)}{2}\right)\\ &=\frac14\left(\frac32-2\cos(2x)+\frac12\cos(4x)\right)\\ &=\frac38-\frac12\cos(2x)+\frac18\cos(4x). \end{aligned} $$

Now integrate term by term. The frequency of \(\cos(2x)\) is \(2\), and the frequency of \(\cos(4x)\) is \(4\):

$$ \begin{aligned} \int \sin^4 x\,dx &=\frac{3x}{8}-\frac12\left(\frac{\sin(2x)}{2}\right) +\frac18\left(\frac{\sin(4x)}{4}\right)+C\\ &=\frac{3x}{8}-\frac{\sin(2x)}{4} +\frac{\sin(4x)}{32}+C. \end{aligned} $$

Differentiate to check both frequency factors and signs:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{3x}{8}-\frac{\sin(2x)}{4} +\frac{\sin(4x)}{32}\right) &=\frac38-\frac{2\cos(2x)}{4}+\frac{4\cos(4x)}{32}\\ &=\frac38-\frac12\cos(2x)+\frac18\cos(4x)\\ &=\sin^4 x. \end{aligned} $$

The final equality uses the expansion obtained above, confirming the result.

Keeping Track of the Constant Term

Every cosine term in the reduced expression has an antiderivative involving sine. A constant, however, does not: \(\int c\,dx=cx\). In these power-reduction expansions, the constant term is part of the integrand and generally makes an essential contribution to the answer.

Here is a useful way to check the process. First, verify that the reduced expression equals the original power. Then integrate that expression term by term. Finally, differentiate the proposed antiderivative and compare the result with the reduced expression—or, equivalently, with the original power. These checks catch different errors: an incorrect expansion, a missing constant term, and a frequency or sign error during integration.

Procedure: Reduce the even sine power to a constant and cosine terms; integrate each term separately; divide each cosine coefficient by its frequency; and differentiate the result to check it.

Worked Example: Integrating a Squared Sine with a Scaled Angle

Worked Example: Integrating \(\sin^2(3x)\)

The same power-reduction identity applies when the angle is \(3x\):

$$ \sin^2(3x)=\frac{1-\cos(6x)}{2} =\frac12-\frac12\cos(6x). $$

Integrate the constant directly. For the cosine term, the inner angle has derivative \(6\), so \(\int\cos(6x)\,dx=\frac{\sin(6x)}{6}\):

$$ \begin{aligned} \int \sin^2(3x)\,dx &=\int\left(\frac12-\frac12\cos(6x)\right)\,dx\\ &=\frac{x}{2}-\frac12\left(\frac{\sin(6x)}{6}\right)+C\\ &=\frac{x}{2}-\frac{\sin(6x)}{12}+C. \end{aligned} $$

Differentiate to verify the result:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{x}{2}-\frac{\sin(6x)}{12}\right) &=\frac12-\frac{6\cos(6x)}{12}\\ &=\frac12-\frac12\cos(6x)\\ &=\sin^2(3x). \end{aligned} $$

The factor \(6\) from differentiating the inner angle cancels the denominator \(6\) introduced during integration.

A Short Derivation of the Integration Rule

The integration step rests on the chain rule. For a nonzero constant \(a\), differentiating \(\sin(ax)\) gives \(a\cos(ax)\). Dividing by \(a\) cancels that factor, so \(\frac{d}{dx}\left(\frac{\sin(ax)}{a}\right)=\cos(ax)\). Thus \(\frac{\sin(ax)}{a}\) is an antiderivative of \(\cos(ax)\). The constant term rule follows from \(\frac{d}{dx}(cx)=c\).

$$ \frac{d}{dx}\left(\frac{\sin(ax)}{a}\right) =\frac{a\cos(ax)}{a} =\cos(ax),\qquad a\ne 0. $$

Because differentiation is linear, the derivative of a sum is the sum of the derivatives. Therefore, once an even sine power has been correctly rewritten as a finite sum of a constant and cosine terms, integrating those terms separately produces an antiderivative of the whole expression. This also explains why the final differentiation check is decisive.

Worked Example: Integrating the Sixth Power

Worked Example: Integrating \(\sin^6 x\)

Start by writing \(\sin^6 x=(\sin^2 x)^3\) and applying power reduction:

$$ \sin^6 x=\frac18\left(1-\cos(2x)\right)^3. $$

Expand the cube and use \(\cos^2(2x)=\frac{1+\cos(4x)}{2}\) and \(\cos^3(2x)=\frac{3\cos(2x)+\cos(6x)}{4}\). The latter identity follows from \(\cos(3\theta)=4\cos^3\theta-3\cos\theta\), with \(\theta=2x\). Substituting gives:

$$ \begin{aligned} \sin^6 x &=\frac18\left(1-3\cos(2x)+3\cos^2(2x)-\cos^3(2x)\right)\\ &=\frac18\left(1-3\cos(2x)+\frac32+\frac32\cos(4x) -\frac34\cos(2x)-\frac14\cos(6x)\right)\\ &=\frac{5}{16}-\frac{15}{32}\cos(2x) +\frac{3}{16}\cos(4x)-\frac{1}{32}\cos(6x). \end{aligned} $$

Integrate the constant and then each cosine term, dividing its coefficient by the cosine's frequency:

$$ \begin{aligned} \int\sin^6 x\,dx &=\frac{5x}{16}-\frac{15}{32}\frac{\sin(2x)}{2} +\frac{3}{16}\frac{\sin(4x)}{4} -\frac{1}{32}\frac{\sin(6x)}{6}+C\\ &=\frac{5x}{16}-\frac{15}{64}\sin(2x) +\frac{3}{64}\sin(4x)-\frac{1}{192}\sin(6x)+C. \end{aligned} $$

Differentiate term by term to check:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{5x}{16}-\frac{15}{64}\sin(2x) +\frac{3}{64}\sin(4x)-\frac{1}{192}\sin(6x)\right) &=\frac{5}{16}-\frac{30}{64}\cos(2x) +\frac{12}{64}\cos(4x)-\frac{6}{192}\cos(6x)\\ &=\frac{5}{16}-\frac{15}{32}\cos(2x) +\frac{3}{16}\cos(4x)-\frac{1}{32}\cos(6x)\\ &=\sin^6 x. \end{aligned} $$

The coefficients and frequencies agree with the reduced expansion, so the antiderivative checks.

A Common Mistake: Losing a Frequency Factor

One common error is to integrate \(\cos(4x)\) as \(\sin(4x)\), forgetting the factor from the chain rule. For example, the following calculation is wrong:

$$ \int \frac18\cos(4x)\,dx \ne \frac18\sin(4x). $$

Differentiating the proposed right-hand side exposes the error: its derivative is \(\frac12\cos(4x)\), not \(\frac18\cos(4x)\). Since \(\frac{d}{dx}\sin(4x)=4\cos(4x)\), the correction is to divide the coefficient by \(4\):

$$ \int \frac18\cos(4x)\,dx =\frac18\cdot\frac{\sin(4x)}{4} =\frac{1}{32}\sin(4x)+C. $$

A separate bookkeeping error is to omit the constant term after reduction. If the expansion includes \(c_0\), its integral \(c_0x\) must appear in the antiderivative. Differentiating the answer will reveal the omission because it will lack that constant contribution.

Summary

To integrate an even power of sine, use power reduction to express it as a constant and cosine terms. Integrate the constant to obtain a term involving \(x\); integrate each cosine term to a sine term, dividing by its frequency. Keep all signs and coefficients, and differentiate the final expression to check it.

Check Your Understanding

Try these questions using power reduction and verify each antiderivative by differentiation.

  1. Rewrite \(\sin^2 x\) as a constant plus a cosine term. What does the constant contribute when you integrate?
  2. What frequency factor must be included when integrating \(\cos(8x)\)?
  3. Find an antiderivative of \(\sin^2(2x)\), showing the reduced expression and the integration step.
  4. In the expansion of \(\sin^4 x\), which term produces an \(x\)-term in the antiderivative?
  5. Differentiate \(\frac{3x}{8}-\frac{\sin(2x)}{4}+\frac{\sin(4x)}{32}\) and identify the trigonometric power it equals.

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