3.1 Integration by parts

By now we have a fairly thorough procedure for how to evaluate many basic integrals. However, although we can integrate ${\displaystyle \int x\text{sin}(x^2)dx}$ by using the substitution, $u=x^2,$ something as simple looking as ${\displaystyle \int x\text{sin}xdx}$ defies us. Many students want to know whether there is a product rule for integration. There isn’t, but there is a technique based on the product rule for differentiation that allows us to exchange one integral for another. We call this technique integration by parts    .

The integration-by-parts formula

If, $h(x)=f(x)g(x),$ then by using the product rule, we obtain $h^{\prime }(x)=f^{\prime }(x)g(x)+g^{\prime }(x)f(x).$ Although at first it may seem counterproductive, let’s now integrate both sides of this equation: ${\displaystyle \int h^{\prime }(x)dx={\displaystyle \int \left(g(x)f^{\prime }(x)+f(x)g^{\prime }(x)\right)dx.}}$

This gives us

Now we solve for ${\displaystyle \int f(x)g^{\prime }(x)dx:}$

By making the substitutions $u=f(x)$ and $v=g(x),$ which in turn make $du=f^{\prime }(x)dx$ and $dv=g^{\prime }(x)dx,$ we have the more compact form

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.

The natural question to ask at this point is: How do we know how to choose $u$ and $dv?$ Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for L ogarithmic Functions, I nverse Trigonometric Functions, A lgebraic Functions, T rigonometric Functions, and E xponential Functions. This mnemonic serves as an aid in determining an appropriate choice for $u.$

The type of function in the integral that appears first in the list should be our first choice of $u.$ For example, if an integral contains a logarithmic function and an algebraic function , we should choose $u$ to be the logarithmic function, because L comes before A in LIATE. The integral in [link] has a trigonometric function $\text{(}\text{sin}x\text{)}$ and an algebraic function $\left(x\right).$ Because A comes before T in LIATE, we chose $u$ to be the algebraic function. When we have chosen $u,$ $dv$ is selected to be the remaining part of the function to be integrated, together with $dx.$