Why the Interval Matters
Odd powers of sine or cosine can behave very differently depending on the interval of integration. On \(0\le x\le \pi/2\), both sine and cosine are nonnegative, so an odd power has a nonnegative integral. Over a full period, however, the function changes sign, and the positive and negative contributions can cancel.
The quarter-period integrals in this tutorial are especially direct: separate one factor of sine or cosine, rewrite the remaining even power using \(\sin^2 x+\cos^2 x=1\), and substitute. The limits are important. For \(u=\cos x\), the limits run in decreasing order, so reversing them introduces a minus sign.
The Substitution for an Odd Sine Power
Let \(n=2m+1\). In \(\sin^{2m+1}x\), keep one factor of \(\sin x\) and write the other \(2m\) factors as \(m\) powers of \(\sin^2x\). Since \(\sin^2x=1-\cos^2x\), the substitution \(u=\cos x\) turns the integrand into a polynomial in \(u\).
At the lower endpoint \(x=0\), \(u=1\); at the upper endpoint \(x=\pi/2\), \(u=0\). Therefore, the substitution gives an integral from \(1\) to \(0\) and a minus sign. Reversing the limits removes that minus sign:
This formula also explains why the integral is nonnegative: on \(0\le u\le1\), the quantity \(1-u^2\) is nonnegative. Expanding the power with the binomial theorem gives an explicit finite sum. Here \(\binom{m}{k}\) is the binomial coefficient, and \(m\) is a nonnegative integer.
The Substitution for an Odd Cosine Power
For an odd power of cosine, keep one factor of \(\cos x\) and rewrite the remaining factors using \(\cos^2x=1-\sin^2x\). Now set \(u=\sin x\). The endpoint values increase from \(0\) to \(1\), and \(du=\cos x\,dx\), so no limit reversal is needed.
The equality of the sine and cosine integrals follows because both substitutions lead to the same integral from \(0\) to \(1\). It is also consistent with interchanging sine and cosine on this interval. The substitution derivation establishes the result directly for every odd positive integer power.
Worked Examples
Worked Example: Evaluate an Odd Sine Power
Evaluate \(\int_0^{\pi/2}\sin^5x\,dx\). Since \(5=2(2)+1\), separate one sine factor and use \(u=\cos x\):
For a check, an antiderivative in \(x\) is \(F(x)=-\cos x+\frac23\cos^3x-\frac15\cos^5x\). Differentiating gives \(F'(x)=\sin x-2\cos^2x\sin x+\cos^4x\sin x=\sin x(1-\cos^2x)^2=\sin^5x\). Also, \(F(\pi/2)=0\) and \(F(0)=-1+\frac23-\frac15=-\frac{8}{15}\). Thus \(F(\pi/2)-F(0)=\frac{8}{15}\), as computed.
Worked Example: Evaluate an Odd Cosine Power
Evaluate \(\int_0^{\pi/2}\cos^7x\,dx\). Set \(u=\sin x\), so \(du=\cos x\,dx\) and \(\cos^6x=(1-u^2)^3\):
To check, differentiate \(G(x)=\sin x-\sin^3x+\frac35\sin^5x-\frac17\sin^7x\). This gives \(G'(x)=\cos x(1-3\sin^2x+3\sin^4x-\sin^6x)=\cos x(1-\sin^2x)^3=\cos^7x\). Since \(G(\pi/2)=1-1+\frac35-\frac17=\frac{16}{35}\) and \(G(0)=0\), the definite integral is \(\frac{16}{35}\).
Worked Example: Compare a Quarter-Period, Half-Period, and Full-Period Integral
For the odd power \(3\), first use the quarter-period result, then compare the sine and cosine integrals over \(0\le x\le\pi\) and the full period \(0\le x\le2\pi\).
The quarter-period sine integral is:
On \([0,\pi]\), sine is nonnegative and the two halves are equal by reflection about \(x=\pi/2\). Hence \(\int_0^\pi\sin^3x\,dx=2(\frac23)=\frac43\). This also checks by the antiderivative \(F(x)=-\cos x+\frac13\cos^3x\): \(F(\pi)=1-\frac13=\frac23\), \(F(0)=-1+\frac13=-\frac23\), and their difference is \(\frac43\).
For cosine, the quarter-period integral is likewise \(\frac23\), but cosine is negative on \((\pi/2,\pi)\). The half-period integral is zero. Indeed, an antiderivative is \(G(x)=\sin x-\frac13\sin^3x\); both \(G(0)\) and \(G(\pi)\) are zero. Finally, both odd powers have zero full-period integral: shifting \(x\) by \(\pi\) changes the sign of sine and cosine, so the contributions from \([0,\pi]\) and \([\pi,2\pi]\) cancel. In particular, the sine integral is \(\frac43-\frac43=0\) over the full period, while the cosine integral is also zero.
Why Full-Period Cancellation Does Not Apply Here
For an odd positive integer \(n\), the identities \(\sin(x+\pi)=-\sin x\) and \(\cos(x+\pi)=-\cos x\) imply \(\sin^n(x+\pi)=-\sin^n x\) and \(\cos^n(x+\pi)=-\cos^n x\). Splitting an integral over \(0\le x\le2\pi\) into two intervals of length \(\pi\) therefore gives equal and opposite contributions.
and the same reasoning gives \(\int_0^{2\pi}\cos^n x\,dx=0\). This cancellation requires including both sign regions over the full period. It does not imply that the quarter-period integrals are zero: sine and cosine are nonnegative throughout \(0\le x\le\pi/2\), and the substitution formula gives their positive values.
The distinction between sine and cosine on the half-period \(0\le x\le\pi\) is also useful. Sine remains nonnegative there, so its odd power has a positive integral. Cosine changes sign at \(\pi/2\), so its odd power cancels on that half-period. A symmetry argument must match the interval and the sign pattern of the function being integrated.
A Common Sign Mistake
With \(u=\cos x\), the limits change from \(x=0,\pi/2\) to \(u=1,0\), and \(du=-\sin x\,dx\). A common incorrect line is:
This loses the minus sign in \(du=-\sin x\,dx\). The correct substitution is \(-\int_1^0(1-u^2)\,du\), which equals \(\int_0^1(1-u^2)\,du=\frac23\). The negative answer is also impossible because the original integrand is nonnegative on the interval. Either preserve the minus sign while keeping the reversed limits, or reverse the limits and remove the minus sign; do not do only one of these operations.
Using the Method Reliably
For a positive odd power on \(0\le x\le\pi/2\), identify the factor that supplies the differential. Save one sine factor and use \(u=\cos x\) for an odd sine power; save one cosine factor and use \(u=\sin x\) for an odd cosine power. Then transform the limits, expand the resulting polynomial if needed, integrate each term, and evaluate at the endpoints.
Write \(\sin^{2m+1}x=(\sin^2x)^m\sin x\), or \(\cos^{2m+1}x=(\cos^2x)^m\cos x\).
Replace \(\sin^2x\) by \(1-\cos^2x\), or \(\cos^2x\) by \(1-\sin^2x\).
Use \(u=\cos x\) for sine powers or \(u=\sin x\) for cosine powers. Account for the negative differential and any reversed limits.
Evaluate the polynomial integral exactly, then check the result by differentiating an antiderivative or by confirming the sign matches the integrand.
This method is valuable because it reduces an odd trigonometric power to ordinary polynomial integration. It also makes the role of the interval visible: the substitution supplies definite limits, while the sign of the original function determines whether cancellation is possible. On \(0\le x\le\pi/2\), there is no negative region to cancel the positive contribution.
Check Your Understanding
Use substitution and keep track of the limits in each question.
- Evaluate \(\int_0^{\pi/2}\sin^3x\,dx\) using \(u=\cos x\).
- Evaluate \(\int_0^{\pi/2}\cos^5x\,dx\) using \(u=\sin x\).
- Explain why the two quarter-period integrals of \(\sin^{2m+1}x\) and \(\cos^{2m+1}x\) are equal.
- Why is \(\int_0^{2\pi}\sin^{2m+1}x\,dx=0\) even though \(\int_0^{\pi/2}\sin^{2m+1}x\,dx\) is positive?
- Identify the missing sign in the incorrect substitution \(\int_0^{\pi/2}\sin^3x\,dx=\int_1^0(1-u^2)\,du\), where \(u=\cos x\).
