Why the Product Rule Leads to a New Integration Identity
When an integrand is a product, it can be difficult to find an antiderivative directly. The product rule offers a way to reorganize such an integral: it expresses the derivative of a product as the sum of two terms. Integrating that sum and rearranging gives an identity connecting the integral of one product to the integral of another.
The key is to keep track of which factor is differentiated and which is treated as an antiderivative. This tutorial derives that relationship from the product rule and checks it in concrete examples.
The Product Rule in the Form We Need
For differentiable functions \(u\) and \(v\), the product rule states that
There are two terms on the right. The term \(u'v\) contains the derivative of \(u\) multiplied by \(v\); the term \(uv'\) contains \(u\) multiplied by the derivative of \(v\). To turn this into an integral identity, integrate both sides over an interval \([a,b]\).
Assume that \(u\) and \(v\) are continuously differentiable on \([a,b]\). Then the product \(uv\) is continuously differentiable there, so the Fundamental Theorem of Calculus applies to \((uv)'\). It also applies to the integrals of \(u'v\) and \(uv'\).
Deriving the Definite Integral Identity
For every \(x\) in the interval, \((uv)'(x)=u'(x)v(x)+u(x)v'(x)\).
Integrating each term from \(a\) to \(b\) gives \(\int_a^b (uv)'(x)\,dx=\int_a^b u'(x)v(x)\,dx+\int_a^b u(x)v'(x)\,dx\).
The left side is \(u(b)v(b)-u(a)v(a)\), also written \([uv]_a^b\).
Subtract \(\int_a^b u'(x)v(x)\,dx\) from both sides.
Here \([uv]_a^b\) means \(u(b)v(b)-u(a)v(a)\). The minus sign appears because the term involving \(u'v\) is subtracted when it is moved to the other side of the equality.
What the Factors Are Doing
The identity has a consistent structure. On the left, \(u\) is multiplied by \(v'\), the derivative of \(v\). On the right, the product \(uv\) appears at the endpoints, and the remaining integral contains \(u'\), the derivative of \(u\), multiplied by \(v\).
In other words, the derivative moves from one factor to the other, while the product \(uv\) supplies a boundary term. The two factors have different roles: \(v'\) is integrated to recover \(v\), while \(u\) is differentiated to produce \(u'\). This description is simply the product rule rearranged; it does not depend on a special choice of functions.
Worked Examples of the Identity
Worked Example: A Polynomial Product on a Finite Interval
Take \(u(x)=x^2\), \(v(x)=x\), and the interval \([0,1]\). Then \(u'(x)=2x\) and \(v'(x)=1\). The integral on the left is
The boundary term is
The remaining integral is
Therefore the right side of the identity is \(1-\frac{2}{3}=\frac{1}{3}\), which agrees with the left side. This example checks both the boundary term and the integral involving \(u'\).
Worked Example: A Polynomial and a Trigonometric Factor
Let \(u(x)=x\), \(v(x)=\sin x\), and use \([0,\pi]\). Then \(u'(x)=1\) and \(v'(x)=\cos x\). The left side is \(\int_0^\pi x\cos x\,dx\). An antiderivative is \(x\sin x+\cos x\), as direct differentiation verifies:
Thus the left side evaluates to
For the right side, the boundary term is
The remaining integral is \(\int_0^\pi \sin x\,dx=[-\cos x]_0^\pi=1-(-1)=2\). Hence the right side is \(0-2=-2\), matching the left side.
Worked Example: Two Trigonometric Factors
Let \(u(x)=\cos x\), \(v(x)=\sin x\), and take the interval \([0,\pi/4]\). Then \(u'(x)=-\sin x\) and \(v'(x)=\cos x\). The left side is \(\int_0^{\pi/4}\cos^2 x\,dx\). Using \(\cos^2 x=(1+\cos 2x)/2\),
The boundary term is
Because \(u'v=(-\sin x)\sin x=-\sin^2 x\), the right side of the identity is the boundary term plus \(\int_0^{\pi/4}\sin^2 x\,dx\). Using \(\sin^2 x=(1-\cos 2x)/2\), this remaining integral is
Consequently the right side is \(\frac{1}{2}+\frac{\pi}{8}-\frac{1}{4}=\frac{\pi}{8}+\frac{1}{4}\), equal to the left side.
How the Identity Relates to Antiderivatives
The definite identity includes a boundary term because integration over \([a,b]\) records the values at both endpoints. There is also a corresponding statement about antiderivatives. Suppose \(u\) and \(v\) are differentiable on an interval and the integrals under discussion have antiderivatives there. If \(B\) is any antiderivative of \(u'v\), then \(B'=u'v\). Differentiate the expression \(uv-B\):
Thus \(uv-B\) is an antiderivative of \(uv'\). Since any two antiderivatives of the same function on an interval differ by a constant, this gives the indefinite-integral relation
where \(C\) is an arbitrary constant. This form is the same rearrangement as the definite identity, but without endpoint evaluation. The constant matters: an indefinite integral represents a family of antiderivatives, not just one function.
A Common Sign Mistake
A frequent error is to use a plus sign in front of the remaining integral. The product rule shows why that cannot be correct. The integral of \(u'v\) is added to the integral of \(uv'\) before rearranging; isolating \(uv'\) requires subtracting the \(u'v\) integral.
For a concrete check, use \(u=x\) and \(v=\sin x\). The incorrect sign would suggest \(x\sin x+\int\sin x\,dx=x\sin x-\cos x+C\) as an antiderivative of \(x\cos x\). Differentiating this proposed answer gives
which is not \(x\cos x\) in general. With the correct minus sign before the remaining integral, the result is \(x\sin x-\int\sin x\,dx=x\sin x+\cos x\), and its derivative is \(x\cos x\), as required.
Why This Derivation Matters
The identity is useful because it changes the form of an integral while preserving equality. One product term is replaced by a boundary product minus another integral. The derivative of one factor and the antiderivative of the other are not arbitrary additions: they are exactly the ingredients produced by the product rule.
The derivation also provides a reliable way to remember the relationship. If the order of the factors or the sign is uncertain, return to \((uv)'=u'v+uv'\), integrate, and isolate the desired term. The identity is therefore both a method for reorganizing integrals and a direct consequence of a familiar differentiation rule.
Check Your Understanding
Use the product rule and the derived identity to answer these questions.
- Starting from \((uv)'=u'v+uv'\), explain why the definite identity contains a minus sign before \(\int_a^b u'v\,dx\).
- For \(u(x)=x^3\) and \(v(x)=x\) on \([0,1]\), evaluate the left side and both terms on the right of the definite identity.
- What does the boundary notation \([uv]_a^b\) mean?
- If \(B'=u'v\), differentiate \(uv-B\) and identify the function whose antiderivative it is.
- What incorrect derivative results from the proposed expression \(x\sin x-\cos x\) for an antiderivative of \(x\cos x\)?
