Section 3.1 Integration by Parts
References.
- Calculus, Early Transcendentals by Stewart, Section 7.1.
Introduction.
The method of Integration by Parts adapts the Product Rule for derivatives in a way similar to how the Chain Rule for derivatives leads to the method of Substitution in Section 1.5. The key fact is the formula:
\begin{equation}
\int f(x) g'(x) \, dx = f(x)g(x) - \int g(x) f'(x) \, dx\tag{3.1.1}
\end{equation}
or with the short-hands \(u = f(x)\) and \(v = g(x)\text{,}\)
\begin{equation}
\int u \frac{dv}{dx} \, dx = u v - \int v \frac{du}{dx} \, dx\tag{3.1.2}
\end{equation}
Along with Substitution, this is one of the main tools that help us to integrate many products of functions.
It is often used in the more concise “differential” form
\begin{equation}
\int u \, dv = u v - \int v \, du\tag{3.1.3}
\end{equation}
derived below.
We will see that this is particularly useful for integrating functions that are products of polynomials, sines, cosines, exponentials, logarithms and inverse trigonometric functions.
A First Exploration.
There are many ways to express a given integral \(\int f(x) dx\) in the product form seen above; let us explore some possibilities, loking for hints about how to choose the factors \(u\) and \(dv/dx\) so that the above formula is not only true but also useful, in that the new integral at right is easier to evaluate.
- First, use \(u=x\) and thus \(dv/dx = \cos x\text{.}\)
- Then try with some other combinations, like
- \(u = \cos x\text{,}\) \(dv/dx = x\)
- \(u = x\cos x\text{,}\) \(dv/dx = 1\)
- \(u = 1\text{,}\) \(dv/dx = x\cos x\)
and think about why the original choice works best. - It is often worthwhile to check an indefinite integral evaluation by differentiating the result, so do this.
Deriving From the Product Rule for Derivatives.
The formula (3.1.1) comes from the Product Rule for differentiation. Starting from \([f(x) g(x)]' = f'(x) g(x) + f(x) g'(x)\) and integrating gives
\begin{equation*}
\int f'(x) g(x) \, dx + \int f(x) g'(x) \, dx = f(x) g(x) + C.
\end{equation*}
Subtracting off the first integral on each side
\begin{equation*}
\int f(x) g'(x) \, dx = f(x) g(x) + C - \int f'(x) g(x) \, dx
\end{equation*}
But the indefinite integral at right already includes an arbitrary constant, so the “\(+C\)” is not needed; discarding that gives (3.1.1) above.
Deriving The Other Forms.
\begin{equation*}
f(x) = u, \, g(x) = v, \text{ so that } f'(x) = du/dx \text{ and } g'(x) = dv/dx
\end{equation*}
\begin{equation*}
du = f'(x) \, dx = (du/dx) dx \text{ and } dv = g'(x) \, dx = (dv/dx) dx
\end{equation*}
and inserting these pieces into either of those equations gives (3.1.3).
Using the Differential Form \(\int u \, dv = uv - \int v \, du\).
A useful strategy for doing Integration by Parts is this:
- Choose a function \(u \; [=f(x)]\) that is a factor of the integrand. (No need to introduce the name \(f(x)\) though; just use \(u\text{.}\))
- Give the name \(dv\) to all the rest of the integral, including the differential \(dx\text{,}\) so that the original integral is now in the form \(\int u \, dv\text{.}\)
-
Integrate the \(dv\) part, to get \(v = \int dv \; [= g(x)]\text{.}\) (Again, no need to introduce the name \(g(x)\text{.}\))Note that no constant of integration is needed: any choice of \(v\) with the correct derivative will do.
- Differentiate \(u=f(x)\) to get a differential expression \(du = f'(x) dx\text{.}\)
- Insert \(u\text{,}\) \(v\) and \(dv\) into Equation (3.1.3) to get an expression for\begin{equation*} \ds uv - \int v \, du, = f(x)g(x) - \int g(x) f'(x)\, dx \end{equation*}
- Note that you still have to evaluate the new integral at right!
Example 3.1.2. Evaluating \(\int x \cos x \, dx,\) again, this time using the differential approach.
- In the first exercise, we could start with choice \(u=x\text{,}\) so that \(dv = \cos x \, dx\text{.}\)
- Then the “partial integration” of \(dv = \cos x \, dx\) gives \(v = \sin x\text{,}\) and differentiation of \(u = x\) gives \(du = dx\text{.}\)
- Thus the new integral to be evaluated is \(-\int v \, du = -\int \sin x \, dx, = \cos x + C\text{,}\) so the full answer is \(u v - \int v \, du = x \sin x + \cos x + C\text{.}\)
Choosing \(u\) (and thus \(dv\)).
How does one choose the part \(u\) to differentiate, which then forces the choice of the part \(dv = g'(x) dx\) to be integrated? Some guidelines are:
- Try to integrate as much as possible in the first integration, which gives \(v\text{,}\) so put as much as possible into \(dv\text{,}\) and as little as possible into the term \(u\) that gets differentiated.
- Try to choose \(u\) so that its derivative is “simpler” than \(u\text{,}\) or at least not more complicated. (For example, positive integer powers of \(x\) are good candidates for \(u\text{.}\))
Sometimes, there is no factor that you know how to integrate, and yet integration by parts can still help, by differentiating everything at the first step. That is, use \(v = x\text{,}\) \(dv = dx\text{,}\) so \(u\) is the whole integrand.
This is useful for integrating the inverses of common functions like inverse trigonometric, inverse hyperbolic, and logarithmic.
Checkpoint 3.1.3. Evaluate \(\ds \int \sinh^{-1} x \, dx\).
Repeated Application.
Sometimes the next integration needed also requires integration by parts; this is useful for example when one factor is a positive integer power of the variable.
Checkpoint 3.1.4. Evaluate \(\ds \int t^2 e^t \, dt\).
At each step, discuss the possible choices for \(u\) and for \(dv\text{.}\)
Note that when repeating, never “backtrack” by integrating the result of differentiation in the previous step and vice verca. Instead, a second stage should use a new \(u\) coming from the derivative of the previous choice of \(u\text{.}\)
Repeated application with a twist.
Finally, sometimes repeated integration by parts almost takes you full circle even if you avoid backtracking, but the solution comes from some equation solving; this happens mainly in the special but important case of exponential-trigonometric products:
Checkpoint 3.1.5. Evaluate \(\ds \int e^x \cos 2x \, dx\).
Section Study Guide.
Study Calculus Volume 2, Section 3.1 2 ; in particular
- Theorems 1 and 2,
- all Examples and Checkpoints,
- one or two exercises from the ranges 1–5 and several from 6–37.
openstax.org/books/calculus-volume-2/pages/3-1-integration-by-partsopenstax.org/books/calculus-volume-2/pages/3-1-integration-by-parts
