Calculus/Integration techniques/Integration by Parts

Integration techniques/Recognizing Derivatives and the Substitution Rule Integration techniques/Trigonometric Substitution Continuing on the path of reversing derivative rules in order to make them useful for integration, we reverse the product rule. Integration by parts Integration by parts If where and are functions of , then Rearranging, Therefore, Therefore, or This is the integration by-parts formula. It is very useful in many integrals involving products of functions and others. For instance, to treat we choose and . With these choices, we have and , and we have Note that the choice of and was critical. Had we chosen the reverse, so that and , the result would have been The resulting integral is no easier to work with than the original; we might say that this application of integration by parts took us in the wrong direction. So the choice is important. One general guideline to help us make that choice is, if possible, to choose as the factor of the integrated, which becomes simpler when we differentiate it. In the last example, we see that does not become simpler when we differentiate it: is no more straightforward than . An important feature of the integration-by-parts method is that we often need to apply it more than once. For instance, to integrate we start by choosing and to get Note that we still have an integral to take care of, and we do this by applying integration by parts again, with and , which gives us So, two applications of integration by parts were necessary, owing to the power of in the integrand. Note that any power of x does become simpler when we differentiate it, so when we see an integral of the form one of our first thoughts ought to be to consider using integration by parts with . Of course, for it to work, we need to be able to write down an antiderivative for . Example Use integration by parts to evaluate the integral Solution: If we let and , then we have and . Using our rule for integration by parts gives We do not seem to have made much progress. But if we integrate by parts again with and and hence and , we obtain We may solve this identity to find the anti-derivative of and obtain With definite integral The rule is essentially the same for definite integrals as long as we keep the endpoints. Integration by parts for definite integrals Suppose f and g are differentiable and their derivatives are continuous. Then . This can also be expressed in Leibniz notation. More Examples Examples Set 1: Integration by Parts Exercises Evaluate the following using integration by parts. 1. 2. 3. 4. 5. Solutions External links Videos and exercises on integration by parts at Khan Academy Integration techniques/Recognizing Derivatives and the Substitution Rule Integration techniques/Trigonometric Substitution