AP BC Calc –4.1 (cont’d cont’d) Integration by Parts and Trigonometric IntegralsFerguson Notes In the first part of 4.1, the part addressing the method of Integration by Substitution, we found that the Substitution Rule was a response to differentiation by the Chain rule. That is, to integrate something that was differentiated by the Chain Rule, we must use Integration by Substitution. But what do we do if we’re faced with something that was differentiated using the Product (or Quotient) Rule? To integrate products (which includes quotients), we use a method called Integration by Parts. Note that the product rule says the following: ( ) ( )( ) ( )( )( )f x g xfx g xf x gx. Therefore, ( ) ( )( )( )( ) ( )fx g xf x gxdxf x g xC. This equation can be rearranged as such: ( )( )( ) ( )( ) ( )f x gx dxf x g xfx g x dx. Is that much better? No? Fine, then let ( )f xuand ( )g xv, which means ( )dufx dxand ( )dvg x dx. By substitution, the equation becomes much easier to see and understand (and remember): udvuvvduIntegration by parts is pretty much that….substitution. The hardest part, in my point of view is choosing the right functions for u and dv…..Ex. 1 –find sinxxdx. At this point, we can see that using the Substitution Rule is going to do us no good in evaluating this integral, so the next best method to try in integration by parts. We also knew that we’d need to integrate this by parts because it is a product of two functions (not always a reliable method of remembering, though). If we let uxand sindvx, then dudxand cosvx . So by utilizing our integration by parts method, sincoscoscoscoscossinxxdxxxxdxxxxdxxxxC . Done. “Well that was easy.”Ex. 2 –find lnxdx. Let lnuxand dvdx, then 1duxand vx. 1lnlnlnlnxdxxxxdxxxdxxxxCx. Wow –these integrals are going quickly!