Chapter 8
Integral Calculus and Its Uses
8.4 Applying Differentiation Rules to Integration
8.4.3 Integration by Parts
We return to the determination of integrals of products. This time we will use the Product Rule for differentiation. For example, suppose we want to evaluate
In Activity 3 you will connect this problem with the Product Rule,
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Find a function \(G(t)\) such that
Let \(F(t)=t\), and write out the Product Rule for this choice of \(F\) and \(G\).
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Solve the resulting equation for \(t\, \sin(t)\).
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Use the result to fill in the blank:
\(t\,\sin(t)= \displaystyle\frac{d}{dt} ( )\). Write your result in part (c) in terms of an indefinite integral.
In Activity 3 the Product Rule enabled us to replace the problem of finding an antiderivative for \(t\,\sin(t)\) with the problem of finding an antiderivative for \(\cos(t)\) - which we already knew how to do. That's what this inverse use of the Product Rule enables us to do - trade in one antidifferentiation problem for another.
Now we need a better notation. For this we turn again to differentials. We rewrite the Product Rule in differential notation with \(f\) representing the derivative of \(F\) and \(g\) representing the derivative of \(G\):
We can rewrite this equation in the form
When we integrate both sides, we obtain
Now, if we write \(u=G(t)\) and \(v=F(t)\), then \(du=g(t) \, dt\) and \(dv = f(t)\,dt\). In this notation, our equation takes the simpler form
This inverse of the Product Rule is called the integration-by-parts formula. When we apply this formula, we first identify the parts \(u\) and \(dv\) - the second of which must be a differential, i.e., must include the factor \(dt\). Then we calculate \(du\) (by differentiating \(u\) and \(v\) (by antidifferentiating \(dv\)) and set up the right-hand side.
Find \(\displaystyle\int\;te^{-3t}\,dt\).
Solution There are two obvious choices for \(u\) and \(dv\):
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\(u=t\) and \(dv = e^{-3t}dt\) or
\(u=e^{-3t}\) and \(dv = t\,dt\).
We will examine both.
If \(u=t\) and \(dv = e^{-3t}dt\), then \(du = dt\), and we may set \(v=-\frac{1}{3} e^{-3t}\). Then
| \(\displaystyle\int\; t \,e^{-3t}\,dt\) | \(\displaystyle= t\,\left(-\frac{1}{3} e^{-3t}\right)- \int\;\left(-\frac{1}{3} e^{-3t}\right)\,dt\) |
| \(\displaystyle=-\frac{1}{3} t\,e^{-3t}+\frac{1}{3} \int\;e^{-3t}\,dt\). |
Since we just found an antiderivative for \(e^{-3t}\), we can use it to evaluate the integral on the right:
Now suppose we set \(u=e^{-3t}\) and \(dv = t\,dt\). Then \(du = -3e^{-3t}\,dt\), and we may set \(v=\frac{t^2}{2}\). For this identification of \(u\) and \(dv\), the integration-by-parts formula becomes
| \(\displaystyle\int\;t\,e^{-3t}\,dt\) | \(\displaystyle= e^{-3t}\,\frac{t^2}{2}-\int\; \frac{t^2}{2} \left(-3e^{-3t}\right)\,dt\) |
| \(\displaystyle=e^{-3t}\, \frac{t^2}{2} + \frac{3}{2} \int\;t^2 \,e^{-3t}\,dt\). |
This formula is valid but not helpful. The new integral on the right is more complicated than the original one
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The integration-by-parts formula allows you to trade in one integration problem for another. There is no guarantee that the new problem is easier than the old. You may need to try several identifications of \(u\) and \(dv\) before you find one that is useful. Ideally, you want \(u\) to be a function whose differential will be simpler than itself and \(dv\) to be a differential whose integral will be simpler than itself. In Example 3, choice (1) has \(du\) simpler than \(u\) and \(v\) of about the same complexity as \(dv\). In choice (2), \(du\) was of the same complexity as \(u\), and \(v\) was more complicated than \(dv\). This last fact is what caused choice (2) to fail.
Use the integration-by-parts formula to evaluate \(\displaystyle\int\,t^2\,\cos\,t\,dt\).
Solution We can simplify the integrand somewhat by reducing the exponent in \(t^2\), so we'll set \(u=t^2\). Then \(du = 2 t\,dt\). Now we must set \(dv = \cos\,t \,dt\), and we may choose \(v=\sin\,t\). With these identifications, the integration-by-parts formula gives
\(\displaystyle\int\;t^2\,\cos\, t\,dt = t^2\,\sin\,t-\int\,2t\,\sin\,t\,dt\).
The remaining integral, apart from a constant factor of \(2\), is the same as the one you evaluated in Activity 3, so we may complete the calculation by using the result from that activity:
| \(\displaystyle\int\;t^2\,\cos\,t\,dt\) | \(\displaystyle = t^2\,\sin\,t-2 \int\;t\,\sin\,t\,dt\) |
| \(=t^2\,\sin\,t-2(-t\,\cos\,t+\sin\,t)+C\) | |
| \(=t^2 \,\sin\,t+2t\,\cos\,t-2\,\sin\,t+C\). |
This is an example of using integration by parts successively, each time reducing the complexity of the integral remaining to be evaluated, until we finally arrive at an integral that can be evaluated from known results.
