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Section6.2Integration by Parts

Here's a simple integral that we can't yet evaluate: \begin{equation*} \int x\cos(x) \,dx. \end{equation*}

It's a simple matter to take the derivative of the integrand using the Product Rule, but there is no Product Rule for integrals. However, this section introduces Integration by Parts, a method of integration that is based on the Product Rule for derivatives. It will enable us to evaluate this integral.

The Product Rule says that if \(u\) and \(v\) are functions of \(x\text{,}\) then \((uv)' = u'v + uv'\text{.}\) For simplicity, we've written \(u\) for \(u(x)\text{,}\) \(u'\) for \(u'(x)\text{,}\) \(v\) for \(v(x)\text{,}\) and \(v'\) for \(v'(x)\text{.}\) Suppose we integrate both sides with respect to \(x\text{.}\) This gives \begin{equation*} \int (uv)'\,dx = \int (u'v+uv')\,dx. \end{equation*}

By the Fundamental Theorem of Calculus, the left side integrates to \(uv\text{.}\) The right side can be broken up into two integrals, and we have \begin{equation*} uv = \int u'v\,dx + \int uv'\,dx. \end{equation*}

Solving for the second integral we have \begin{equation*} \int uv'\,dx = uv - \int u'v\,dx. \end{equation*}

Using differential notation, we can write \(du = u'(x)dx\) and \(dv=v'(x)dx\) and the expression above can be written as follows: \begin{equation*} \int u\,dv = uv - \int v\,du. \end{equation*}

This is the Integration by Parts formula. For reference purposes, we state this in a theorem.

The integration by parts formula can also be written as \begin{equation*} \int f(x)\ g'(x) \ dx=f(x)g(x)-\int \fp(x)\ g(x)\ dx \end{equation*} for differentiable functions \(f\) and \(g\text{.}\)

Let's try an example to understand our new technique.

Example6.2.2Integrating using Integration by Parts

Evaluate \(\ds\int x\cos(x) \ dx\text{.}\)


You may wonder what would have happened in Example 6.2.2 if we had chosen our \(u\) and \(dv\) differently. If we had chosen \(u=\cos(x)\) and \(dv=x \ dx\) then \(du=-\sin(x)\ dx\) and \(v=x^2/2\text{.}\) Our second integral is not simpler than the first, we would have \(\int x\cos(x)\,dx=\cos(x)\frac{x^2}{2}-\int \frac{x^2}{2}\left(-\sin(x)\right)\,dx\text{.}\) The only way to approach this second integral would be yet another integration by parts.

Example 6.2.2 demonstrates how Integration by Parts works in general. We try to identify \(u\) and \(dv\) in the integral we are given, and the key is that we usually want to choose \(u\) and \(dv\) so that \(du\) is simpler than \(u\) and \(v\) is hopefully not too much more complicated than \(dv\text{.}\) This will mean that the integral on the right side of the Integration by Parts formula, \(\int v\,du\) will be simpler to integrate than the original integral \(\int u\,dv\text{.}\)

In the example above, we chose \(u=x\) and \(dv=\cos(x) \,dx\text{.}\) Then \(du=dx\) was simpler than \(u\) and \(v=\sin(x)\) is no more complicated than \(dv\text{.}\) Therefore, instead of integrating \(x\cos(x) \,dx\text{,}\) we could integrate \(\sin(x) \,dx\text{,}\) which we knew how to do.

A useful mnemonic for helping to determine \(u\) is “LIATE,” where

L = Logarithmic, I = Inverse Trig., A = Algebraic (polynomials, roots, power functions),T = Trigonometric, and E = Exponential.

If the integrand contains both a logarithmic and an algebraic term, in general letting \(u\) be the logarithmic term works best, as indicated by L coming before A in LIATE.

We now consider another example.

Example6.2.4Integrating using Integration by Parts

Evaluate \(\displaystyle \int x e^x\,dx\text{.}\)

Example6.2.6Integrating using Integration by Parts

Evaluate \(\displaystyle \int x^2\cos(x) \,dx\text{.}\)

Example6.2.9Integrating using Integration by Parts

Evaluate \(\displaystyle \int e^x\cos(x) \,dx\text{.}\)

Example6.2.12Integrating using Integration by Parts: antiderivative of \(\ln(x)\)

Evaluate \(\displaystyle \int \ln(x) \,dx\text{.}\)

Example6.2.14Integrating using Int. by Parts: antiderivative of \(\arctan x\)

Evaluate \(\displaystyle \int \arctan x \,dx\text{.}\)


Subsection6.2.1Substitution Before Integration

When taking derivatives, it was common to employ multiple rules (such as using both the Quotient and the Chain Rules). It should then come as no surprise that some integrals are best evaluated by combining integration techniques. In particular, here we illustrate making an “unusual” substitution first before using Integration by Parts.

Example6.2.15Integration by Parts after substitution

Evaluate \(\ds \int \cos(\ln(x) )\ dx\text{.}\)


Subsection6.2.2Definite Integrals and Integration By Parts

So far we have focused only on evaluating indefinite integrals. Of course, we can use Integration by Parts to evaluate definite integrals as well, as Theorem 6.2.1 states. We do so in the next example.

Example6.2.16Definite integration using Integration by Parts

Evaluate \(\displaystyle \int_1^2 x^2 \ln(x) \,dx\text{.}\)


In general, Integration by Parts is useful for integrating certain products of functions, like \(\int x e^x\,dx\) or \(\int x^3\sin(x) \,dx\text{.}\) It is also useful for integrals involving logarithms and inverse trigonometric functions.

As stated before, integration is generally more difficult than derivation. We are developing tools for handling a large array of integrals, and experience will tell us when one tool is preferable/necessary over another. For instance, consider the three similar–looking integrals \begin{equation*} \int xe^x\,dx, \qquad \int x e^{x^2}\,dx \qquad \text{ and } \qquad \int xe^{x^3}\,dx. \end{equation*}

While the first is calculated easily with Integration by Parts, the second is best approached with Substitution. Taking things one step further, the third integral has no answer in terms of elementary functions, so none of the methods we learn in calculus will get us the exact answer.

Integration by Parts is a very useful method, second only to substitution. In the following sections of this chapter, we continue to learn other integration techniques. Section 6.3 focuses on handling integrals containing trigonometric functions.