6.1: Integration by Parts

We've seen integrals that can be evaluated by a substitution. For example, we evaluate \(\int \cos 2x \di x\) by substituting \(u = 2x.\) In this section we build on this knowledge by exploring how to reexpress integrals as simpler expressions. Because we use the Product Rule to differentiate a product of functions, let's reverse this process by learning to integrate a product of functions.

Let \(u\) and \(v\) be two differentiable functions of \(x.\) By the Product Rule, the derivative of the product \(uv\) is \[(uv)' = u' v + u v' \pd \] Integrating both sides gives \[ \ba \int (uv)' \di x &= \int u' v \di x + \int u v' \di x \nl uv &= \int u' v \di x + \int u v' \di x \nl \int u v' \di x &= uv - \int u' v \di x \pd \ea \] We condense the terms \(u' \di x\) and \(v' \di x\) as \(\dd u\) and \(\dd v,\) respectively. Thus, we have \begin{equation} \int u \di v = uv - \int v \di u \label{eq:ibp} \pd \end{equation} In words, \(\eqref{eq:ibp}\) allows us to reexpress an integral \(\int u \di v\) by choosing a term \(u\) to differentiate and an expression \(\dd v\) to integrate. The resulting integral, \(\int v \di u,\) is hopefully easier to evaluate. We call this procedure Integration by Parts.

TIP Try to let \(u\) be a function whose derivative is simpler than \(u.\) Hence, when we select \(u\) we often prioritize logarithmic functions or inverse trigonometric functions. Of course, exceptions apply; no set of rules can replace trial and error.

REMARK In Example 1 our choices of \(u\) and \(\dd v\) succeeded because they yielded the simpler integral \(\int e^x \di x.\) But flipping our choices—that is, selecting \(u = e^x\) and \(\dd v = x \di x\) (from which \(\dd u = e^x \di x\) and \(v = \tfrac{1}{2} x^2\))—would instead give \[\int x e^x \di x = \tfrac{1}{2} x^2 e^x - \int \tfrac{1}{2} x^2 e^x \di x \pd\] While this equation is true, it does not help us since the integral \(\int \frac{1}{2} x^2 e^x \di x\) is more complicated to evaluate.

For this reason, it is insufficient to simply memorize \(\eqref{eq:ibp};\) you must strategically choose \(u\) and \(\dd v.\) Integration is a skill; the choices of \(u\) and \(\dd v\) are not always obvious. With continued practice, you will gain an intuition for selecting each component.

Because Integration by Parts enables us to reexpress an integral, we are not performing a substitution. The quantities \(u\) and \(\dd v\) are simply placeholders; they do not initiate a variable change. So in Example 4, we did not change the bounds of the integral. Just as we use a combination of differentiation rules, Integration by Parts can be paired with other integration techniques—most notably, the Substitution Rule, as shown by the following example.

The graph of \(y = \asin x\) is positive in the first quadrant, so the area is given by the definite integral \[\int_0^1 \asin x \di x \pd\] Figure 1 shows this bounded region \(R.\)

Antidifferentiation To find \(\int \asin x \di x,\) let's conduct Integration by Parts with \[ \baat{2} u &= \asin x \lspace &\dd v &= \dd x \nl \dd u &= \frac{1}{\sqrt{1 - x^2}} \di x \lspace &v &= x \pd \eaat \] So \(\eqref{eq:ibp}\) gives \[\int \asin x \di x = x \asin x - \int \frac{x}{\sqrt{1 - x^2}} \di x \pd\] We evaluate \(\int x/\sqrt{1 - x^2} \di x\) by substituting \(w = 1 - x^2.\) Then \(\dd w = -2x \di x\) and so the integral becomes \[-\int \frac{1}{2\sqrt w} \di w = -\tfrac{1}{2} \par{2 \sqrt w} + C = -\sqrt{1 - x^2} + C \pd\] So we have \[ \ba \int \asin x \di x &= x \asin x - \par{-\sqrt{1 - x^2}} + C \nl &= x \asin x + \sqrt{1 - x^2} + C \pd \ea \]

Area By the Fundamental Theorem of Calculus, the area is \[ \ba \int_0^1 \asin x \di x &= \par{x \asin x + \sqrt{1 - x^2}} \intEval_0^1 \nl &= \par{\asin 1 + 0} - \par{\sqrt 1} \nl &= \boxed{\frac{\pi}{2} - 1} \ea \]