Recognize when to use integration by parts compared to other integration methods
Use the integration by parts formula to solve indefinite integrals
Apply integration by parts to evaluate definite integrals
Integration-by-Parts
You’ve already learned how to tackle many basic integrals, and you can handle something like [latex]\displaystyle\int x\sin\left({x}^{2}\right)dx[/latex] using substitution with [latex]u={x}^{2}[/latex]. But what about [latex]\displaystyle\int x\sin{x}dx[/latex]? This seemingly simple integral requires a different approach.
Many students wonder if there’s a product rule for integration—there isn’t. However, we can use a technique based on the product rule for differentiation to exchange one integral for another. This powerful method is called integration by parts.
The key insight is that when you have a product of two functions, you can often rewrite the integral in a more manageable form. Let’s see how this works by deriving the integration by parts formula.
If, [latex]h\left(x\right)=f\left(x\right)g\left(x\right)[/latex], then by using the product rule, we obtain [latex]{h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right)[/latex]. Although at first it may seem counterproductive, let’s now integrate both sides of this equation: [latex]\displaystyle\int {h}^{\prime }\left(x\right)dx=\displaystyle\int \left(g\left(x\right){f}^{\prime }\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)\right)dx[/latex].
By making the substitutions [latex]u=f\left(x\right)[/latex] and [latex]v=g\left(x\right)[/latex], which in turn make [latex]du={f}^{\prime }\left(x\right)dx[/latex] and [latex]dv={g}^{\prime }\left(x\right)dx[/latex], we have the more compact form
Let [latex]u=f\left(x\right)[/latex] and [latex]v=g\left(x\right)[/latex] be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:
The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.
Use integration by parts with [latex]u=x[/latex] and [latex]dv=\sin{x}dx[/latex] to evaluate [latex]\displaystyle\int x\sin{x}dx[/latex].
By choosing [latex]u=x[/latex], we have [latex]du=1dx[/latex]. Since [latex]dv=\sin{x}dx[/latex], we get [latex]v=\displaystyle\int \sin{x}dx=\text{-}\cos{x}[/latex]. It is handy to keep track of these values as follows:
At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen [latex]u=\sin{x}[/latex] and [latex]dv=x[/latex]. If we had done so, then we would have [latex]du=\cos{x}[/latex] and [latex]v=\frac{1}{2}{x}^{2}[/latex]. Thus, after applying integration by parts, we have [latex]{\displaystyle\int }^{\text{ }}x\sin{x}dx=\frac{1}{2}{x}^{2}\sin{x}-{\displaystyle\int }^{\text{ }}\frac{1}{2}{x}^{2}\cos{x}dx[/latex]. Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for [latex]u[/latex] and [latex]dv[/latex] before finding a choice that works.
Second, you may wonder why, when we find [latex]v={\displaystyle\int }^{\text{ }}\sin{x}dx=\text{-}\cos{x}[/latex], we do not use [latex]v=\text{-}\cos{x}+K[/latex]. To see that it makes no difference, we can rework the problem using [latex]v=\text{-}\cos{x}+K\text{:}[/latex]