Why shifted limits matter
A definite integral can be easier to evaluate after its interval is translated to a familiar one, such as \([0,\pi]\) or \([0,2\pi]\). For sine-cosine powers, however, translating the limits by \(\pi\) can reverse the sign of the integrand. A shift by \(2\pi\) does not.
The sign depends on the combined powers of sine and cosine. Once that dependence is clear, a translated integral can be related to an integral on a simpler interval before any integration is done.
What a shift by pi does
For any integer \(k\), the angle-addition identities give \(\sin(x+k\pi)=(-1)^k\sin x\) and \(\cos(x+k\pi)=(-1)^k\cos x\). Raising these expressions to the powers \(m\) and \(n\), respectively, shows that the product changes by the factor \((-1)^{k(m+n)}\).
In particular, shifting by \(\pi\) preserves the product when \(m+n\) is even and reverses its sign when \(m+n\) is odd. Shifting by \(2\pi\) preserves the product for either parity.
Why the translation rule works
The integrand identity explains how the function changes, but the limits must be translated consistently as well. The substitution \(x=t+k\pi\) moves the shifted interval back to the original one. Its differential is \(dx=dt\), so it introduces no additional factor.
In the integral over \([a+k\pi,b+k\pi]\), set \(x=t+k\pi\). The limits for \(t\) become \(a\) and \(b\).
Use \(\sin(t+k\pi)=(-1)^k\sin t\) and \(\cos(t+k\pi)=(-1)^k\cos t\). The product gains the factor \((-1)^{k(m+n)}\).
Since \(dx=dt\), the result is \((-1)^{k(m+n)}\int_a^b\sin^m t\cos^n t\,dt\), which is the stated rule.
This argument applies to positive and negative integers \(k\). It also shows why both endpoints must move together: the substitution translates the whole interval, not just one of its bounds.
Worked examples
Worked Example: An odd total power shifted by pi
Evaluate \(\int_{\pi}^{2\pi}\sin^3 x\cos^2 x\,dx\) by translating it to \([0,\pi]\).
The total power is \(3+2=5\), which is odd. The interval \([\pi,2\pi]\) is \([0,\pi]\) shifted by \(k=1\), so the translation factor is \((-1)^{1\cdot5}=-1\). First evaluate the integral on the simpler interval:
Set \(u=\cos t\), so \(du=-\sin t\,dt\). At \(t=0\), \(u=1\); at \(t=\pi\), \(u=-1\). Therefore
Apply the shift factor to get the requested value:
Check: Set \(x=t+\pi\) directly. Then \(\sin(t+\pi)=-\sin t\) and \(\cos(t+\pi)=-\cos t\), so \(\sin^3(t+\pi)\cos^2(t+\pi)=-\sin^3t\cos^2t\). Integrating for \(0\le t\le\pi\) gives \(-4/15\), as found.
Worked Example: An odd total power shifted by two pi
Evaluate \(\int_{2\pi}^{3\pi}\sin x\cos^2 x\,dx\). The total power is \(1+2=3\), but the shift is \(2\pi\), so the value is unchanged. Translate to \([0,\pi]\):
With \(u=\cos t\), \(du=-\sin t\,dt\), and the limits \(u=1\) at \(t=0\) and \(u=-1\) at \(t=\pi\), this becomes
Thus the shifted integral is \(2/3\). Check: Writing \(x=t+2\pi\) leaves both sine and cosine unchanged, so the integrand and the translated limits give exactly the integral on \([0,\pi]\).
Worked Example: An even total power shifted by pi
Evaluate \(\int_{\pi}^{2\pi}\sin^2 x\,dx\). Here \(m+n=2\), so shifting by \(\pi\) does not change the value. Translate the interval to \([0,\pi]\), then use the power-reduction identity \(\sin^2 t=(1-\cos 2t)/2\):
Check: Since \(\sin(t+\pi)=-\sin t\), squaring gives \(\sin^2(t+\pi)=\sin^2t\). The translated integral is therefore the integral over \([0,\pi]\), consistent with the calculation.
A reliable way to handle shifted limits
When an interval is presented as a familiar interval shifted by a multiple of \(\pi\), separate the translation from the integration. First identify the shift \(k\pi\) and translate both endpoints back. Then determine whether \(k(m+n)\) is even or odd. Only after applying the resulting factor should you evaluate the integral on the simpler interval.
Express the given limits as \(a+k\pi\) and \(b+k\pi\), using the same integer \(k\).
Add the sine and cosine exponents to find \(m+n\).
Use \((-1)^{k(m+n)}\): an even exponent gives \(+1\), and an odd exponent gives \(-1\).
Use an appropriate integration method for that interval, then include the sign factor.
The translation rule does not evaluate the integral on the simpler interval by itself. It only relates the two integrals. The integration method still depends on the powers and limits, as in the earlier material on sine and cosine powers.
A common mistake: checking only the sine power
A frequent error is to decide the sign from the sine exponent alone. For example, it would be incorrect to claim that \(\sin^2x\cos x\) is unchanged by a shift of \(\pi\) just because the sine power is even. The cosine factor also changes sign.
The total power is \(2+1=3\), which is odd, so the sign reverses. The correction is to count all sine and cosine factors together: the shift factor is determined by \(m+n\), not by either exponent in isolation.
Another essential detail is to shift both limits by the same amount. The formula relates \([a+k\pi,b+k\pi]\) to \([a,b]\); it does not say that changing just the upper or lower limit creates a translated copy of the original integral.
When the rule is useful
Many sine-cosine power integrals are given over intervals whose endpoints are not the most convenient ones for direct evaluation. If both endpoints are displaced by a multiple of \(\pi\), the translation formula can move the calculation to a simpler interval. It can also reveal whether two integrals have the same value or opposite values without evaluating either one.
The distinction between a shift by \(\pi\) and one by \(2\pi\) is especially important. An odd total power reverses sign under a single \(\pi\)-shift, but a \(2\pi\)-shift preserves the value. An even total power preserves the value under either shift. The total power and the amount of translation together determine the sign.
Check Your Understanding
Use the translation rule before attempting to evaluate each integral.
- Does \(\int_{2\pi}^{3\pi}\sin^4x\cos x\,dx\) equal, or have the opposite sign from, \(\int_0^\pi\sin^4x\cos x\,dx\)?
- What factor relates \(\int_{-\pi}^{0}\sin^2x\cos^4x\,dx\) to \(\int_0^\pi\sin^2x\cos^4x\,dx\)?
- Why does shifting an interval by \(2\pi\) leave a sine-cosine power integral unchanged, regardless of the exponents?
- For \(\sin^3x\cos^2x\), what factor relates the integral over \([3\pi,4\pi]\) to the integral over \([0,\pi]\)?
- What must be true of the two endpoints for the translation formula to apply directly?
