Integrating Even Powers of Cosine

Why Even Powers Need a Different Approach

For an odd power of cosine, one factor of cosine can often be kept separate while the remaining even power is rewritten using \(\sin^2 x+\cos^2 x=1\). An even power has no such leftover factor. Instead, the power-reduction identity converts it into a sum of terms involving cosines of multiple angles.

This approach is useful because each resulting term has a direct antiderivative. The main care required is to keep track of the frequencies: integrating \(\cos(kx)\) introduces a factor of \(1/k\).

Definition: An even power of cosine has the form \(\cos^{2n}(x)\), where \(n\) is a nonnegative integer. To integrate it by power reduction, repeatedly use \(\cos^2 x=(1+\cos 2x)/2\), simplify the resulting expression, and integrate its constant and cosine terms.

The Identity and the Integration Rule

The power-reduction identity follows from the double-angle formula \(\cos(2x)=2\cos^2 x-1\). Solving this equation for \(\cos^2 x\) gives:

\cos^2 x=\frac{1+\cos(2x)}{2}.

When applying the identity to a higher even power, it is useful to first view that power as a product of factors of \(\cos^2 x\). For instance, \(\cos^4 x=(\cos^2 x)^2\). Expand the resulting expression, and apply power reduction again to any even power that remains.

After expansion, the integral is a sum of a constant and terms of the form \(\cos(kx)\). For a nonzero constant \(k\), the chain rule gives the integration rule:

\int \cos(kx)\,dx=\frac{\sin(kx)}{k}+C.

Indeed, differentiating \(\sin(kx)/k\) gives \(\cos(kx)\). A constant term \(a\) integrates to \(ax\), since the derivative of \(ax\) is \(a\). These rules apply to each term of a finite sum by the linearity of integration.

A Base Case: The Square

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

Apply power reduction directly:

\begin{aligned} \int \cos^2 x\,dx &=\int \frac{1+\cos(2x)}{2}\,dx\\ &=\frac{1}{2}\int 1\,dx+\frac{1}{2}\int \cos(2x)\,dx\\ &=\frac{x}{2}+\frac{1}{2}\left(\frac{\sin(2x)}{2}\right)+C\\ &=\frac{x}{2}+\frac{\sin(2x)}{4}+C. \end{aligned}

To check, differentiate the proposed antiderivative:

\frac{d}{dx}\left(\frac{x}{2}+\frac{\sin(2x)}{4}\right) =\frac{1}{2}+\frac{2\cos(2x)}{4} =\frac{1+\cos(2x)}{2} =\cos^2 x.

The last equality is the power-reduction identity, so the derivative matches the integrand.

Applying Power Reduction More Than Once

For a fourth power, the first application of the identity leaves a term involving \(\cos^2(2x)\). That term is still an even power of cosine, so reduce it again. This is the essential repeated step: do not stop while an even power remains in the expanded expression.

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

Write the fourth power as the square of \(\cos^2 x\), then apply the identity:

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

Apply power reduction to the remaining square, using the angle \(2x\):

\cos^2(2x)=\frac{1+\cos(4x)}{2}.

Substitute and collect the constant and cosine terms:

\begin{aligned} \cos^4 x &=\frac{1+2\cos(2x)+\frac{1+\cos(4x)}{2}}{4}\\ &=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x). \end{aligned}

Now integrate term by term. The frequencies are \(2\) and \(4\), so their antiderivatives include factors \(1/2\) and \(1/4\), respectively:

\begin{aligned} \int \cos^4 x\,dx &=\int\left(\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)\right)\,dx\\ &=\frac{3x}{8}+\frac{1}{2}\left(\frac{\sin(2x)}{2}\right) +\frac{1}{8}\left(\frac{\sin(4x)}{4}\right)+C\\ &=\frac{3x}{8}+\frac{\sin(2x)}{4}+\frac{\sin(4x)}{32}+C. \end{aligned}

Check by differentiation:

\begin{aligned} \frac{d}{dx}\left(\frac{3x}{8}+\frac{\sin(2x)}{4}+\frac{\sin(4x)}{32}\right) &=\frac{3}{8}+\frac{2\cos(2x)}{4}+\frac{4\cos(4x)}{32}\\ &=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)\\ &=\cos^4 x. \end{aligned}

The final equality uses the expansion obtained by repeated power reduction.

More Terms from a Higher Even Power

As the exponent increases, the expansion produces more multiple-angle terms. Some of these can themselves be even powers, which must be reduced before integrating. In the next example, one remaining term is \(\cos^3(2x)\). This is not an even power, but it can be rewritten using the triple-angle identity \(\cos(3\theta)=4\cos^3\theta-3\cos\theta\). Solving that identity for \(\cos^3\theta\) gives \(\cos^3\theta=(3\cos\theta+\cos(3\theta))/4\).

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

First write the sixth power as a cube of \(\cos^2 x\), then expand:

\begin{aligned} \cos^6 x &=(\cos^2 x)^3\\ &=\left(\frac{1+\cos(2x)}{2}\right)^3\\ &=\frac{1+3\cos(2x)+3\cos^2(2x)+\cos^3(2x)}{8}. \end{aligned}

Reduce the square using power reduction and rewrite the cube using the triple-angle identity with \(\theta=2x\):

\cos^2(2x)=\frac{1+\cos(4x)}{2}, \qquad \cos^3(2x)=\frac{3\cos(2x)+\cos(6x)}{4}.

Substituting these expressions gives the following coefficients. The constant terms inside the outer factor \(1/8\) are \(1+3/2=5/2\). The \(\cos(2x)\) terms are \(3+3/4=15/4\). Therefore:

\begin{aligned} \cos^6 x &=\frac{1}{8}\left( \frac{5}{2}+\frac{15}{4}\cos(2x)+\frac{3}{2}\cos(4x)+\frac{1}{4}\cos(6x) \right)\\ &=\frac{5}{16}+\frac{15}{32}\cos(2x)+\frac{3}{16}\cos(4x)+\frac{1}{32}\cos(6x). \end{aligned}

Integrate each term, dividing each sine term by its frequency:

\begin{aligned} \int \cos^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\sin(2x)}{64} +\frac{3\sin(4x)}{64}+\frac{\sin(6x)}{192}+C. \end{aligned}

Differentiate to verify all the terms:

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

The derivative matches the expanded form of the integrand, confirming the antiderivative.

Why the Method Works

The method rests on two facts: power reduction gives an exact identity, and integration is linear. The following steps show why these facts produce a valid antiderivative.

Replace a square exactly.
The identity \(\cos^2 u=(1+\cos(2u))/2\) holds for every real \(u\), so it can be applied not only to \(x\), but also to angles such as \(2x\) or \(3x\).

Expand and collect terms.
Algebra rewrites the original power as a finite sum of constants and cosine terms. Each equality in the expansion preserves the value of the integrand.

Integrate each term.
A constant integrates to a multiple of \(x\), and \(\cos(kx)\) integrates to \(\sin(kx)/k\) when \(k\ne 0\). Linearity allows these antiderivatives to be added.

Check by differentiation.
Differentiating the sum recovers each term of the expanded integrand. Since the expansion equals the original power, the derivative is the original integrand.

This argument also explains the role of the frequency factor. By the chain rule, the derivative of \(\sin(kx)\) is \(k\cos(kx)\). Dividing by \(k\) compensates for that extra factor. The rule requires \(k\ne 0\); the constant term is integrated separately.

A Common Frequency Error

A frequent mistake is to integrate a multiple-angle cosine as though its angle were just \(x\). For example, after power reduction a term \(\frac{1}{2}\cos(2x)\) may appear. Omitting the frequency factor gives an incorrect result:

\int \frac{1}{2}\cos(2x)\,dx \ne \frac{1}{2}\sin(2x).

The proposed expression on the right differentiates to \(\cos(2x)\), not \(\frac{1}{2}\cos(2x)\). The correct calculation includes division by the frequency \(2\):

\int \frac{1}{2}\cos(2x)\,dx =\frac{1}{2}\cdot\frac{\sin(2x)}{2} =\frac{\sin(2x)}{4}.

Always check each sine term by differentiating it. This quickly reveals a missing frequency factor or an incorrect coefficient.

When to Stop Reducing

Continue reducing while an even power remains in the expanded expression. Once the expression consists only of a constant and individual cosine terms, integrate directly. For a base case, \(\cos^0 x=1\), so its integral is \(x+C\). For a positive even exponent, the examples show how the same identity can be applied at the original angle and at the multiple angles created during expansion.

The result may contain a term proportional to \(x\), coming from the constant part of the expansion, as well as sine terms at multiple angles. These are all part of the antiderivative. The additive constant \(C\) accounts for the fact that any two antiderivatives of the same function differ by a constant.

Check Your Understanding

Use power reduction and check your results by differentiation.

Rewrite \(\cos^2 x\) as a constant plus a multiple-angle cosine.
In integrating \(\cos(5x)\), what factor must divide \(\sin(5x)\), and why?
Reduce \(\cos^2(3x)\) to a constant and a cosine term.
What frequency error occurs if \(\frac{1}{2}\cos(2x)\) is integrated as \(\frac{1}{2}\sin(2x)\)?
After expanding a power-reduction expression, what should you check before integrating term by term?

Integrating Even Powers of Cosine

What You'll Learn

Why Even Powers Need a Different Approach

The Identity and the Integration Rule