Why does this mnemonic work? Remember that whatever we pick to be d v d v must be something we can integrate. When we have chosen u, u, d v d v is selected to be the remaining part of the function to be integrated, together with d x. Because A comes before T in LIATE, we chose u u to be the algebraic function. The integral in Example 3.1 has a trigonometric function ( sin x ) ( sin x ) and an algebraic function ( x ). For example, if an integral contains a logarithmic function and an algebraic function, we should choose u u to be the logarithmic function, because L comes before A in LIATE. The type of function in the integral that appears first in the list should be our first choice of u. This mnemonic serves as an aid in determining an appropriate choice for u. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. The natural question to ask at this point is: How do we know how to choose u u and d v ? d v ? 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. Įvaluate ∫ x e 2 x d x ∫ x e 2 x d x using the integration-by-parts formula with u = x u = x and d v = e 2 x d x. ∫ h ′ ( x ) d x = ∫ ( g ( x ) f ′ ( x ) + f ( x ) g ′ ( x ) ) d x. Although at first it may seem counterproductive, let’s now integrate both sides of this equation: ∫ h ′ ( x ) d x = ∫ ( g ( x ) f ′ ( x ) + f ( x ) g ′ ( x ) ) d x. h ′ ( x ) = f ′ ( x ) g ( x ) + g ′ ( x ) f ( x ). If, h ( x ) = f ( x ) g ( x ), h ( x ) = f ( x ) g ( x ), then by using the product rule, we obtain h ′ ( x ) = f ′ ( x ) g ( x ) + g ′ ( x ) f ( x ). We call this technique integration by parts. There isn’t, but there is a technique based on the product rule for differentiation that allows us to exchange one integral for another. Many students want to know whether there is a product rule for integration. However, although we can integrate ∫ x sin ( x 2 ) d x ∫ x sin ( x 2 ) d x by using the substitution, u = x 2, u = x 2, something as simple looking as ∫ x sin x d x ∫ x sin x d x defies us. 3.1.3 Use the integration-by-parts formula for definite integrals.īy now we have a fairly thorough procedure for how to evaluate many basic integrals.3.1.2 Use the integration-by-parts formula to solve integration problems.3.1.1 Recognize when to use integration by parts.
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