From Antiderivatives to Definite Integrals
For an indefinite integral, the main task is to find an antiderivative. For a definite integral, there is one more step: evaluate that antiderivative at both endpoints and subtract in the correct order. This is especially important for products of an exponential and a sine or cosine, because both the exponential value and the trigonometric phase can change from one endpoint to the other.
Let \(a,b,\phi\) be real numbers and let \(u\leq v\) be finite real endpoints. When \(a\) and \(b\) are not both zero, the antiderivatives needed here are
and
These formulas include unshifted products by taking \(\phi=0\). The focus here is their definite-integral use: substitute the upper endpoint, substitute the lower endpoint, and subtract the second value from the first.
General Endpoint Formulas
Writing the endpoint subtraction explicitly helps keep the phase and signs visible. For \(a^2+b^2>0\), the sine integral is
For the cosine integral, the corresponding formula is
In each expression, the entire lower-endpoint term is subtracted. Parentheses matter: the minus sign applies to every part of that term.
Why the Endpoint Formulas Work
The endpoint step follows from the Fundamental Theorem of Calculus. To justify that the displayed expressions are suitable antiderivatives, differentiate them. Put \(\theta=bx+\phi\) and \(D=a^2+b^2\). For the sine antiderivative, the product rule gives
Here \(\theta'=b\), so the derivative of \(a\sin\theta-b\cos\theta\) is \(ab\cos\theta+b^2\sin\theta\). For the cosine antiderivative, differentiation gives
Both integrands are continuous on every finite interval. The Fundamental Theorem therefore gives their definite integrals by subtracting the antiderivative value at \(u\) from its value at \(v\). This also explains why the endpoint order is not optional: reversing it changes the sign of the integral.
Worked Examples
Worked Example: An Unshifted Exponential-Sine Integral
Evaluate \(\int_0^\pi e^x\sin x\,dx\).
Here \(a=1\), \(b=1\), and \(\phi=0\), so \(a^2+b^2=2\). An antiderivative is
At the upper endpoint, \(\sin\pi=0\) and \(\cos\pi=-1\), giving
At the lower endpoint, \(\sin 0=0\) and \(\cos 0=1\), so
Subtract the lower value from the upper value:
Check by differentiating the proposed antiderivative. The sine-antiderivative calculation above, with \(a=b=1\) and \(\phi=0\), gives \(F'(x)=e^x\sin x\). The endpoint values are \(F(\pi)=e^\pi/2\) and \(F(0)=-1/2\), so their difference reproduces the stated result.
Worked Example: An Exponential-Cosine Integral
Evaluate \(\int_0^{\pi/2}e^{2x}\cos(3x)\,dx\).
Now \(a=2\), \(b=3\), and \(\phi=0\), so \(a^2+b^2=4+9=13\). Use the cosine antiderivative:
At \(x=\pi/2\), the angle is \(3\pi/2\), where \(\cos(3\pi/2)=0\) and \(\sin(3\pi/2)=-1\). Thus
At \(x=0\), \(\cos 0=1\) and \(\sin 0=0\), hence
Therefore,
For a derivative check, set \(\theta=3x\). Differentiating \(G\) gives
The evaluated endpoint difference is therefore the required integral.
Worked Example: A Shifted Phase in a Sine Integral
Evaluate \(\int_0^{\pi/2}e^x\sin(2x+\pi/6)\,dx\).
The phase is \(2x+\pi/6\), so the shift \(\pi/6\) must be included at both endpoints. Here \(a=1\), \(b=2\), and \(a^2+b^2=5\). An antiderivative is
At \(x=\pi/2\), the phase is \(\pi+\pi/6=7\pi/6\). Since \(\sin(7\pi/6)=-1/2\) and \(\cos(7\pi/6)=-\sqrt{3}/2\),
At \(x=0\), the phase is \(\pi/6\), with sine \(1/2\) and cosine \(\sqrt3/2\). Therefore,
Subtracting gives
Differentiate to check: the general sine derivative identity with \(a=1\), \(b=2\), and \(\phi=\pi/6\) gives \(F'(x)=e^x\sin(2x+\pi/6)\). The endpoint values just calculated differ by the displayed answer, confirming both the phase substitution and the endpoint arithmetic.
Zero Parameters and Basic Conditions
The denominator \(a^2+b^2\) is positive unless \(a=b=0\). If exactly one parameter is zero, the formulas remain valid. For example, if \(b=0\), then \(\sin(bx+\phi)=\sin\phi\) is constant and the sine formula reduces to the exponential antiderivative when \(a\ne0\). If \(a=0\) and \(b\ne0\), the formulas reduce to the usual antiderivatives of a trigonometric function, with the phase retained.
If \(a=b=0\), the integrands are constants: \(e^{0x}\sin(\phi)=\sin\phi\) and \(e^{0x}\cos(\phi)=\cos\phi\). In that case the definite integrals are simply
There is no singularity in the integral itself; only the antiderivative formulas with denominator \(a^2+b^2\) are inapplicable when both parameters vanish.
Phase Shifts and Endpoint Accuracy
A phase shift changes the angle at which the sine or cosine is evaluated; it does not change the endpoint values of \(x\). For an integral on \([u,v]\), evaluate the phase as \(bu+\phi\) at the lower endpoint and \(bv+\phi\) at the upper endpoint. The exponential is evaluated separately as \(e^{au}\) and \(e^{av}\).
It is possible to change variables to make the trigonometric phase the new variable, but that also changes the exponential factor and the interval. Using the antiderivative directly usually keeps the bookkeeping simpler: substitute each original \(x\)-endpoint into the entire antiderivative.
A Common Sign Mistake
In the first example, a common error is to subtract in the wrong order and write \(F(0)-F(\pi)\). That gives
This is the negative of the correct integral. The correction is to apply the endpoint rule as \(F(\pi)-F(0)\), which gives \((e^\pi+1)/2\). A related error is to subtract only part of the lower-endpoint expression. Writing \(F(v)-\bigl(F(u)\bigr)\) with parentheses makes clear that the whole lower value is subtracted.
Why These Integrals Are Useful
Exponential-sine and exponential-cosine products occur whenever an exponential weight is combined with oscillation. Their definite integrals summarize the total signed accumulation over a specified interval. The endpoint formulas make that accumulation explicit, while also showing how changing the interval or phase changes the result.
The same procedure applies to any finite interval, not only intervals whose endpoints are familiar trigonometric angles. In such cases, the endpoint values may remain in terms of sine and cosine. Exact simplification is helpful when available, but it is not required for the endpoint rule to apply.
Summary
- For \(a^2+b^2>0\), use the exponential-sine or exponential-cosine antiderivative with the same phase as the integrand.
- Evaluate the full antiderivative at both endpoints and calculate \(F(v)-F(u)\).
- For a shifted phase, substitute \(bu+\phi\) and \(bv+\phi\) at the respective endpoints.
- When \(a=b=0\), integrate the resulting constant directly.
- Check the antiderivative by differentiation and guard against reversing the endpoint order.
Check Your Understanding
Use the antiderivative formulas and endpoint rule to answer these questions. Check each result by differentiating the antiderivative you use.
- Evaluate \(\int_0^{\pi/2}e^x\sin x\,dx\).
- For \(\int_0^\pi e^{-x}\cos(2x+\pi/3)\,dx\), what phase is used at each endpoint?
- State the endpoint formula for \(\int_u^v e^{ax}\cos(bx+\phi)\,dx\) when \(a^2+b^2>0\).
- What is \(\int_1^4 e^{0x}\sin(\pi/6)\,dx\), and why do the exponential-trigonometric antiderivative formulas with denominator \(a^2+b^2\) not apply?
- If an endpoint calculation produces \(F(u)-F(v)\) instead of \(F(v)-F(u)\), how is the result related to the desired integral?
