Integrating Odd Powers of Sine

Why Odd Powers of Sine Are Convenient

An integral containing a power of sine may look difficult because the power is attached to a trigonometric function. When the exponent is odd, however, one factor of sine can be kept separate. The other factors can be rewritten using the identity $\sin^2(x)=1-\cos^2(x)$. The remaining sine factor is exactly what is needed for a substitution involving cosine.

This method applies to an integral of $\sin^m(x)$ when $m$ is a positive odd integer. Write the exponent as $m=2k+1$, where $k$ is a nonnegative integer. Then $\sin^{2k+1}(x)$ contains $2k$ factors that can be paired into sine squares, along with one sine factor left over.

Definition: A positive odd power of sine is an integrand of the form $\sin^m(x)$, where $m=2k+1$ for some integer $k\geq 0$. Its basic integration method is to write $\sin^{2k+1}(x)=\bigl(\sin^2(x)\bigr)^k\sin(x)$, replace each $\sin^2(x)$ with $1-\cos^2(x)$, and substitute $u=\cos(x)$.

The method is a direct use of substitution and the Pythagorean identity. It does not require a recurrence: after rewriting the integrand, the integral becomes a polynomial integral in $u$.

The Substitution, Including Its Sign

For an odd exponent $2k+1$, first separate one sine factor:

\sin^{2k+1}(x) = \bigl(\sin^2(x)\bigr)^k\sin(x) = \bigl(1-\cos^2(x)\bigr)^k\sin(x).

Now set $u=\cos(x)$. Differentiating this substitution gives $du=-\sin(x)\,dx$, or equivalently $\sin(x)\,dx=-du$. The minus sign matters: the leftover sine factor is replaced by $-du$, not $du$. Thus, for an indefinite integral,

\int \sin^{2k+1}(x)\,dx = -\int (1-u^2)^k\,du.

The power $(1-u^2)^k$ is a polynomial because $k$ is a nonnegative integer. Expand it, integrate each term, and then replace $u$ with $\cos(x)$. An arbitrary constant of integration must be included.

Key takeaway: For every nonnegative integer $k$, the integral of $\sin^{2k+1}(x)$ reduces to integrating the polynomial $-(1-u^2)^k$ with respect to $u$, where $u=\cos(x)$.

A General Formula and Why It Works

The binomial theorem expands $(1-u^2)^k$ as a finite sum. In particular,

(1-u^2)^k = \sum_{j=0}^{k}(-1)^j\binom{k}{j}u^{2j}.

Integrating each power $u^{2j}$ gives $u^{2j+1}/(2j+1)$. Including the minus sign from $du=-\sin(x)\,dx$, and substituting $u=\cos(x)$, yields the general formula below.

Theorem: If $k$ is a nonnegative integer, then $$ \int \sin^{2k+1}(x)\,dx = -\sum_{j=0}^{k}(-1)^j\binom{k}{j} \frac{\cos^{2j+1}(x)}{2j+1}+C. $$ Here $C$ is an arbitrary constant. The formula follows by separating one sine factor, substituting $u=\cos(x)$, expanding the resulting polynomial, and integrating term by term.

The formula can also be justified by differentiation, which checks the result without relying on a memorized pattern. Differentiate the expression on the right. The derivative of $\cos^{2j+1}(x)/(2j+1)$ is $-\cos^{2j}(x)\sin(x)$, so differentiating the whole sum gives

\begin{aligned} \frac{d}{dx}\left( -\sum_{j=0}^{k}(-1)^j\binom{k}{j} \frac{\cos^{2j+1}(x)}{2j+1} \right) &= \sin(x)\sum_{j=0}^{k}(-1)^j\binom{k}{j}\cos^{2j}(x)\\ &= \sin(x)\bigl(1-\cos^2(x)\bigr)^k\\ &= \sin(x)\bigl(\sin^2(x)\bigr)^k\\ &= \sin^{2k+1}(x). \end{aligned}

The second equality uses the binomial theorem; the third uses $\sin^2(x)+\cos^2(x)=1$. This confirms that the displayed expression is an antiderivative on every interval where sine and cosine are defined, which is all real $x$.

Worked Examples

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

Separate one sine factor and rewrite the remaining square:

\begin{aligned} \int \sin^3(x)\,dx &=\int \sin^2(x)\sin(x)\,dx\\ &=\int \bigl(1-\cos^2(x)\bigr)\sin(x)\,dx. \end{aligned}

Let $u=\cos(x)$, so $du=-\sin(x)\,dx$. Therefore,

\begin{aligned} \int \bigl(1-\cos^2(x)\bigr)\sin(x)\,dx &=-\int(1-u^2)\,du\\ &=-\int(1-u^2)\,du\\ &=-\left(u-\frac{u^3}{3}\right)+C\\ &=-u+\frac{u^3}{3}+C\\ &=-\cos(x)+\frac{\cos^3(x)}{3}+C. \end{aligned}

To check, differentiate the proposed antiderivative:

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

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

Worked Example: Integrating $\sin^5(x)$

Write the fifth power as four factors grouped into squares and one remaining sine:

\begin{aligned} \int \sin^5(x)\,dx &=\int \bigl(\sin^2(x)\bigr)^2\sin(x)\,dx\\ &=\int \bigl(1-\cos^2(x)\bigr)^2\sin(x)\,dx. \end{aligned}

Use $u=\cos(x)$ and $du=-\sin(x)\,dx$. Expand the polynomial before integrating:

\begin{aligned} \int \bigl(1-\cos^2(x)\bigr)^2\sin(x)\,dx &=-\int(1-u^2)^2\,du\\ &=-\int(1-2u^2+u^4)\,du\\ &=-\left(u-\frac{2u^3}{3}+\frac{u^5}{5}\right)+C\\ &=-u+\frac{2u^3}{3}-\frac{u^5}{5}+C\\ &=-\cos(x)+\frac{2\cos^3(x)}{3}-\frac{\cos^5(x)}{5}+C. \end{aligned}

Differentiate to check all three terms:

\begin{aligned} \frac{d}{dx}\left(-\cos(x)+\frac{2\cos^3(x)}{3}-\frac{\cos^5(x)}{5}\right) &=\sin(x)-2\cos^2(x)\sin(x)+\cos^4(x)\sin(x)\\ &=\sin(x)\bigl(1-2\cos^2(x)+\cos^4(x)\bigr)\\ &=\sin(x)\bigl(1-\cos^2(x)\bigr)^2\\ &=\sin(x)\sin^4(x)\\ &=\sin^5(x). \end{aligned}

This recovers the original integrand, as required.

Worked Example: Evaluating a Definite Integral of $\sin^5(x)$

The same substitution evaluates definite integrals, but the bounds must change when the variable changes. Consider $\int_0^{\pi/2}\sin^5(x)\,dx$. With $u=\cos(x)$, the lower $x$-bound $0$ gives $u=\cos(0)=1$, and the upper bound $\pi/2$ gives $u=\cos(\pi/2)=0$. Therefore,

\begin{aligned} \int_0^{\pi/2}\sin^5(x)\,dx &=-\int_1^0(1-u^2)^2\,du\\ &=\int_0^1(1-2u^2+u^4)\,du\\ &=\left[u-\frac{2u^3}{3}+\frac{u^5}{5}\right]_0^1\\ &=1-\frac{2}{3}+\frac{1}{5}\\ &=\frac{15}{15}-\frac{10}{15}+\frac{3}{15}\\ &=\frac{8}{15}. \end{aligned}

The result can be checked using the antiderivative from the previous example. At $x=\pi/2$, its value is $0$. At $x=0$, its value is $-1+2/3-1/5=-8/15$. The difference is $0-(-8/15)=8/15$, agreeing with the substitution calculation.

A Reliable Sequence of Steps

For any positive odd power, the work follows the same pattern. Keeping the substitution and its sign visible helps prevent common errors.

Write the exponent as $2k+1$.
Since the exponent is odd, $k$ is a nonnegative integer, and one sine factor can be left unpaired.

Separate the leftover factor.
Write $\sin^{2k+1}(x)=(\sin^2(x))^k\sin(x)$.

Convert the paired factors.
Replace $(\sin^2(x))^k$ with $(1-\cos^2(x))^k$.

Substitute and integrate.
Set $u=\cos(x)$, use $\sin(x)\,dx=-du$, expand the polynomial if needed, and integrate its terms.

Return to $x$ and verify.
Replace $u$ with $\cos(x)$, include $C$ for an indefinite integral, and differentiate the result to check it.

Common Mistakes and How to Correct Them

A frequent error is to rewrite every factor of sine as $1-\cos^2(x)$, forgetting that the identity applies to $\sin^2(x)$, not to $\sin(x)$ itself. For instance, the following is not a valid transformation:

\int \sin^3(x)\,dx \ne \int (1-\cos^2(x))\,dx.

The right-hand integrand has lost a sine factor. It also cannot be integrated by simply treating $\cos(x)$ as the variable while keeping $dx$: when $u=\cos(x)$, the differential relation is $du=-\sin(x)\,dx$. The correct setup preserves that factor:

\int \sin^3(x)\,dx = \int (1-\cos^2(x))\sin(x)\,dx = -\int(1-u^2)\,du.

Another error is dropping the minus sign in $du=-\sin(x)\,dx$. Omitting it reverses the sign of the proposed antiderivative. Differentiating the final expression is an efficient way to catch this: if its derivative is $-\sin^{2k+1}(x)$, the sign has been reversed.

When This Method Applies

The key condition is that the exponent of sine is a positive odd integer. If the exponent is $1$, the method still works with $k=0$: it gives $\int\sin(x)\,dx=-\cos(x)+C$. For exponents $3,5,7,\ldots$, the polynomial $(1-u^2)^k$ has increasingly many terms, but the procedure is unchanged.

This discussion concerns a single sine power, $\sin^m(x)$. In an integral that also contains a power of cosine, the available factors and the useful substitution may depend on both exponents; that is a different choice of setup. Here, the decisive feature is that an odd sine power leaves exactly one factor of $\sin(x)\,dx$ for the substitution $u=\cos(x)$.

Practical check: After integrating, differentiate the proposed answer. The result should simplify to $\sin^{2k+1}(x)$. In particular, keeping the factors in the form $\sin(x)(1-\cos^2(x))^k$ makes the final verification transparent.

Check Your Understanding

Use the odd-power method, and check each indefinite integral by differentiation.

Evaluate $\int\sin^7(x)\,dx$.
In the substitution $u=\cos(x)$, what expression replaces $\sin(x)\,dx$?
For an exponent $m=9$, what is the value of $k$ in $m=2k+1$, and what polynomial must be integrated after substitution?
Evaluate $\int_0^{\pi/2}\sin^3(x)\,dx$, changing the bounds when you substitute.
Explain why rewriting $\sin^3(x)$ as $1-\cos^2(x)$ alone does not preserve the integrand.

What You'll Learn

Why Odd Powers of Sine Are Convenient

The Substitution, Including Its Sign

A General Formula and Why It Works

Worked Examples

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

Worked Example: Integrating \(\sin^5(x)\)

Worked Example: Evaluating a Definite Integral of \(\sin^5(x)\)

A Reliable Sequence of Steps

Common Mistakes and How to Correct Them

When This Method Applies

Check Your Understanding

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Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Integrating Odd Powers of Sine

What You'll Learn

Why Odd Powers of Sine Are Convenient

The Substitution, Including Its Sign

A General Formula and Why It Works

Worked Examples

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

Worked Example: Integrating \(\sin^5(x)\)

Worked Example: Evaluating a Definite Integral of \(\sin^5(x)\)

A Reliable Sequence of Steps

Common Mistakes and How to Correct Them

When This Method Applies

Check Your Understanding

Test yourself: Calculus II final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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