Integration by Parts, calculus lesson

Substitution undoes the chain rule. The product rule needs its own reverse, and that is integration by parts. It handles integrands like $x e^x$ , $x \cos x$ and $\ln x$ , where one factor gets simpler when you differentiate it and the other is easy to integrate.

Key terms

Product rule derivation

Definition Let $u$ and $v$ be differentiable on an interval with $u'$ and $v'$ continuous there. The product rule says $(uv)' = u'v + uv'$ . Rearranging and integrating both sides gives

\int u(x) \, v'(x) \, dx = u(x) \, v(x) - \int u'(x) \, v(x) \, dx.

In plain words An integral of a product can be traded for a different integral of a product. The trade pays off when the new integral is easier than the old one.

Example For $\int x \cos x \, dx$ take $u = x$ and $v' = \cos x$ , so $u' = 1$ and $v = \sin x$ .

\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C.

Differentiating gives $\sin x + x \cos x - \sin x = x \cos x$ .

Definite and indefinite formulas

Definition With $du = u'(x) \, dx$ and $dv = v'(x) \, dx$ , the indefinite formula is written

\int u \, dv = uv - \int v \, du.

If $u'$ and $v'$ are continuous on $[a, b]$ , the definite formula is

\int_a^b u(x) \, v'(x) \, dx = \Big[ u(x) \, v(x) \Big]_a^b - \int_a^b u'(x) \, v(x) \, dx.

A useful guide for choosing $u$ is to pick the factor that becomes simpler when differentiated, such as a power of $x$ or a logarithm, and let $dv$ be the rest, which you must be able to integrate.

In plain words Pick one factor to differentiate and one to integrate. The answer is $u$ times $v$ , minus the integral of $v$ times the derivative of $u$ .

Example For $\int \ln x \, dx$ on $x > 0$ , take $u = \ln x$ and $dv = dx$ , so $du = \frac{1}{x} dx$ and $v = x$ .

\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C.

Boundary terms

Definition In the definite formula the expression

\Big[ u v \Big]_a^b = u(b)v(b) - u(a)v(a)

is the boundary term. It is a number, computed from the values at the endpoints, and it must be evaluated alongside the remaining integral.

In plain words In a definite integral, the $uv$ part gets plugged in at the bounds right away. Often one end gives zero, which saves work.

Example For $\int_0^1 x e^x \, dx$ take $u = x$ , $dv = e^x \, dx$ , so $du = dx$ and $v = e^x$ .

\int_0^1 x e^x \, dx = \Big[ x e^x \Big]_0^1 - \int_0^1 e^x \, dx = (e - 0) - (e - 1) = 1.

The boundary term at $x = 0$ vanished because of the factor $x$ .

Repeated integration by parts

Definition If the new integral $\int v \, du$ is again a product of the same kind, integration by parts can be applied to it as well, each time with the same hypotheses on the new $u$ and $v$ . Two patterns are common. A polynomial factor of degree $k$ disappears after $k$ steps. With $e^{ax}$ times $\sin bx$ or $\cos bx$ , two steps bring back the original integral, and you solve for it algebraically.

In plain words Keep going while each step makes the problem simpler, and stop as soon as the remaining integral is one you know.

Example For $\int x^2 e^x \, dx$ , first take $u = x^2$ , $dv = e^x \, dx$ .

\int x^2 e^x \, dx = x^2 e^x - 2 \int x e^x \, dx.

Then use $\int x e^x \, dx = x e^x - e^x + C$ , from $u = x$ and $dv = e^x \, dx$ .

\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C = e^x (x^2 - 2x + 2) + C.

Reduction formulas

Definition A reduction formula expresses an integral that depends on an integer $n$ through the same kind of integral with a smaller $n$ . For integers $n \ge 2$ , integration by parts with $u = \sin^{n-1} x$ and $dv = \sin x \, dx$ , followed by $\cos^2 x = 1 - \sin^2 x$ , gives

\int \sin^n x \, dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n - 1}{n} \int \sin^{n-2} x \, dx.

For the definite integrals $I_n = \int_0^{\pi/2} \sin^n x \, dx$ the boundary term is zero at both ends, so

I_n = \frac{n - 1}{n} I_{n-2} \quad (n \ge 2), \qquad I_0 = \frac{\pi}{2}, \quad I_1 = 1.

In plain words Instead of fighting a high power directly, you lower the power by two at a time until you reach a case you can do at sight.

Example Two steps of the recursion bring $I_4$ down to $I_0$ ,

I_4 = \frac{3}{4} I_2 = \frac{3}{4} \cdot \frac{1}{2} I_0 = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \approx 0.589.

Check yourself

Find $\int x e^{2x} \, dx$ .

Show the answer

Take $u = x$ and $dv = e^{2x} \, dx$ , so $v = \frac{1}{2} e^{2x}$ . The result is

\int x e^{2x} \, dx = \frac{x}{2} e^{2x} - \frac{1}{2} \int e^{2x} \, dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C.

Find $\int e^x \sin x \, dx$ .

Show the answer

Call it $J$ . With $u = \sin x$ and $dv = e^x \, dx$ we get

J = e^x \sin x - \int e^x \cos x \, dx.

With $u = \cos x$ and $dv = e^x \, dx$ the last integral is $e^x \cos x + J$ . So $J = e^x \sin x - e^x \cos x - J$ , and solving for $J$ gives

J = \frac{e^x}{2}(\sin x - \cos x) + C.

Evaluate $\int_1^e \ln x \, dx$ .

Show the answer

With the antiderivative $x \ln x - x$ found above,

\int_1^e \ln x \, dx = \Big[ x \ln x - x \Big]_1^e = (e - e) - (0 - 1) = 1.

Integration by Parts

Theory, part 109

Practice, part 110

Key terms

Product rule derivation

Definite and indefinite formulas

Boundary terms

Repeated integration by parts

Reduction formulas

Check yourself

How this lesson connects

Builds on

Used later in