Integration by Parts

From the Product Rule to a New Integration Tool

Calculus I gave you one integration technique that undoes composition: substitution, the chain rule read backward. But many integrands are products of unrelated functions — x cos x, x² eˣ, ln x by itself — where no inner function's derivative is loitering nearby waiting to cancel. For these, you need a technique that undoes the product rule. Start from the product rule for differentiation: d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Integrate both sides with respect to x: u(x)v(x) = ∫u'(x)v(x) dx + ∫u(x)v'(x) dx. Rearranging, and writing du = u'(x) dx and dv = v'(x) dx for brevity, gives the integration by parts formula: ∫u dv = uv − ∫v du. You have traded one integral for another — the hope, and the entire skill, is that the new integral ∫v du is easier than the one you started with.

Choosing u and dv

Every factor in the integrand must be assigned to either u or dv. A useful ordering of preference for u, remembered by the acronym LIATE, is: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Pick u from whichever of these categories appears first in the integrand — that factor tends to simplify (or at least not worsen) when differentiated, while dv should be a piece you can actually antidifferentiate. Get the split backward and the new integral is often harder than the original, a reliable warning sign to try again.

Worked Example: ∫x cos x dx

1. By LIATE, the algebraic factor x outranks the trigonometric factor, so set u = x and dv = cos x dx.

2. Then du = dx, and antidifferentiating dv gives v = sin x.

3. Apply the formula: ∫x cos x dx = uv − ∫v du = x sin x − ∫sin x dx.

4. The remaining integral is elementary: ∫sin x dx = −cos x. So the answer is x sin x − (−cos x) + C = x sin x + cos x + C.

Check by differentiating: d/dx[x sin x + cos x] = sin x + x cos x − sin x = x cos x, exactly the original integrand.

Repeated and Cyclic Applications

Some integrands need the formula applied more than once (an algebraic factor raised to a power, differentiated repeatedly until it becomes a constant), and a special family of integrands — a product of an exponential and a sine or cosine — never simplifies to zero no matter how many times you differentiate. That family is solved by a clever algebraic trick worth seeing once.

Worked Example: ∫eˣ cos x dx, the Cyclic Case

1. Call the unknown integral I. Set u = eˣ, dv = cos x dx, so du = eˣ dx and v = sin x. Then I = eˣ sin x − ∫eˣ sin x dx.

2. Attack ∫eˣ sin x dx the same way: u = eˣ, dv = sin x dx, so du = eˣ dx and v = −cos x. This gives ∫eˣ sin x dx = −eˣ cos x + ∫eˣ cos x dx = −eˣ cos x + I.

3. Substitute back: I = eˣ sin x − (−eˣ cos x + I) = eˣ sin x + eˣ cos x − I.

4. The unknown I now appears on both sides. Add I to both sides: 2I = eˣ sin x + eˣ cos x, so I = (eˣ/2)(sin x + cos x) + C.

Nothing here was a limit or an infinite process — it is ordinary algebra applied to an equation that happens to contain the very integral you are solving for. The trick generalizes to any exponential-times-sinusoid integral.

Why This Matters

Integration by parts is the single most-used technique beyond substitution: it evaluates ∫ln x dx, every moment computation in probability (∫x·f(x) dx for a density f), and the reduction formulas that make computer algebra systems work. It is also the engine behind deriving several results you will meet later this course, including the general power series coefficients built from repeated differentiation.

Curriculum aligned with OpenStax's Calculus Volume 2; all lesson text is original to Syllabus.