Why a Phase Shift Matters
A constant inside a trigonometric function changes the angle at which the function is evaluated. In an integral such as \(\int \sin^2(x+a)\,dx\), the expression is not a new kind of sine function: it is the usual sine function evaluated at a translated angle. Recognizing this translation lets us use familiar identities and antiderivatives.
For an indefinite integral, the translation is usually short: set \(u=x+a\), so \(du=dx\). For a definite integral, the same change also translates the endpoints. This second part is essential. The new integrand and the new limits must describe the same interval as the original integral.
Translation of a Shifted Integral
Let \(g\) be a function for which the relevant integral exists, and let \(a\) be a constant. The substitution \(u=x+a\) gives the general definite-integral rule
The direction of the interval is preserved: adding the same constant to both endpoints does not change which endpoint comes first. If the original limits are reversed, the translated limits remain reversed as well.
For an indefinite integral, the corresponding statement is that if \(G'(u)=g(u)\), then
where \(C\) is an arbitrary constant. The chain rule confirms the formula because the derivative of \(G(x+a)\) is \(G'(x+a)\cdot 1=g(x+a)\).
Why the Endpoint Rule Works
The substitution rule can be justified directly using an antiderivative. Suppose \(G'(u)=g(u)\). By the chain rule, \(G(x+a)\) is an antiderivative of \(g(x+a)\). The Fundamental Theorem of Calculus therefore gives
Applying the same theorem to \(g(u)\) on the translated interval gives
Both integrals equal the same difference, which proves the translation rule. The limits are not an optional adjustment: they are what make the second integral have exactly the same antiderivative values as the first.
Applying Familiar Power Identities
Once the angle has been translated, the usual sine-cosine identities can be applied to the new variable. For example, the power-reduction identities for squares are
These identities work just as well when \(u=x+a\). A shifted square does not require a new power-reduction formula; it requires careful handling of the shifted angle and, for definite integrals, the shifted bounds. Likewise, products such as \(\sin^m(x+a)\cos^n(x+a)\) can be expressed in terms of \(u\) and handled using the identities and power-integral techniques already established.
Worked Example: An Indefinite Integral of a Shifted Odd Power
Evaluate \(\int \sin^3(x+\pi/3)\,dx\).
Set \(u=x+\pi/3\). Then \(du=dx\), so the integral becomes
Write one sine factor separately and use \(\sin^2u=1-\cos^2u\):
Let \(v=\cos u\), so \(dv=-\sin u\,du\). Thus
Substitute \(v=\cos u\), then \(u=x+\pi/3\):
Check by differentiation. The derivative of the first term is \(\sin(x+\pi/3)\). The derivative of the second term is \(-\cos^2(x+\pi/3)\sin(x+\pi/3)\). Their sum is \(\sin(x+\pi/3)(1-\cos^2(x+\pi/3))=\sin^3(x+\pi/3)\), as required.
Worked Example: A Definite Integral with Shifted Limits
Evaluate \(\int_0^{\pi/2}\cos^2(x-\pi/6)\,dx\).
Set \(u=x-\pi/6\), so \(du=dx\). Translate both endpoints:
Now apply the cosine-square identity and integrate:
At the upper endpoint, the antiderivative is \(\pi/6+\sqrt{3}/8\). At the lower endpoint, it is \(-\pi/12-\sqrt{3}/8\). Therefore,
For a check, differentiate \(F(u)=u/2+\sin(2u)/4\). Its derivative is \(1/2+\cos(2u)/2=\cos^2u\). Since \(u=x-\pi/6\) has derivative \(1\), \(F(x-\pi/6)\) differentiates to the original integrand. The endpoint evaluation above uses the correct translated interval.
Using Angle-Addition Identities Instead
A phase shift can also be removed by expanding the shifted sine or cosine. The angle-addition identities are
These identities are especially useful when expanding a shifted expression turns it into powers or products of the unshifted sine and cosine. For a power, expansion can create several terms, so it is worth keeping the expression organized and checking that all coefficients are carried through.
Worked Example: Expanding a Shifted Square
Evaluate \(\int_0^{\pi/2}\sin^2(x+\pi/6)\,dx\) by the angle-addition identity.
Since \(\cos(\pi/6)=\sqrt{3}/2\) and \(\sin(\pi/6)=1/2\),
Squaring gives
Integrate each term over the stated interval. The square identities give \(\int_0^{\pi/2}\sin^2x\,dx=\pi/4\) and \(\int_0^{\pi/2}\cos^2x\,dx=\pi/4\). Also, \(\sin x\cos x\) has antiderivative \(\sin^2x/2\), so its integral on this interval is \(1/2\). Hence
To check the antiderivative used in this calculation, an antiderivative of the expanded integrand is
Differentiating the first two terms gives \((3/4)\sin^2x+(1/4)\cos^2x\). Differentiating the last gives \((\sqrt{3}/2)\sin x\cos x\). Their sum is exactly the expanded integrand. Evaluating \(F(\pi/2)-F(0)\) gives \(\pi/4+\sqrt{3}/4\), matching the integral calculation.
The substitution method gives the same value with less expansion: set \(u=x+\pi/6\), which changes the limits from \(0,\pi/2\) to \(\pi/6,2\pi/3\), then integrate \(\sin^2u\). The expansion above illustrates that the angle-addition route is also valid when its terms are integrated correctly.
A Common Endpoint Error
Consider again \(\int_0^{\pi/2}\cos^2(x-\pi/6)\,dx\). A common mistake is to set \(u=x-\pi/6\) in the integrand but retain the old limits \(0\) and \(\pi/2\). That incorrect calculation gives
which is not the original integral's value. The shift changes the interval: its endpoints are \(-\pi/6\) and \(\pi/3\), not \(0\) and \(\pi/2\). The correct evaluation is \(\pi/4+\sqrt{3}/4\). The error is not in the cosine-square identity; it is in pairing the transformed integrand with limits that belong to a different interval.
Choosing Between the Two Representations
Both shifting the variable and expanding with angle addition are exact. Shifting is often direct when every trigonometric factor has the same shifted angle, because the whole expression becomes a familiar function of \(u\). Expansion is useful when the resulting unshifted terms are straightforward to integrate or when the angle-addition identity itself is the clearest way to expose the expression.
For a product with a shared phase, such as \(\sin^m(x+a)\cos^n(x+a)\), the substitution \(u=x+a\) moves both factors together: the product becomes \(\sin^m u\cos^n u\). If the integral is definite, both bounds move by \(a\). If the phases differ between factors, one cannot replace both angles by the same \(u\) without further rewriting; angle-addition identities can express the factors relative to a common variable instead.
In every case, keep the original limits visible until they have been translated, and check an antiderivative by differentiating it with the chain rule. These checks distinguish a correct phase shift from an expression that merely looks similar.
Check Your Understanding
For each question, pay particular attention to what happens to the angle and, where relevant, the endpoints.
- Under the substitution \(u=x-\pi/4\), what are the new limits when \(x\) ranges from \(-\pi/6\) to \(\pi/3\)?
- State an antiderivative of \(\sin^2(x+\pi/3)\) using a power-reduction identity.
- What is the main error in replacing \(\int_0^{\pi/2}g(x+\pi/5)\,dx\) with \(\int_0^{\pi/2}g(u)\,du\)?
- Write \(\cos(x+\pi/3)\) in terms of \(\sin x\) and \(\cos x\).
- How can differentiation check an antiderivative of \(g(x+a)\)?
