8Techniques of IntegrationOver the next few sections we examine some techniques that are frequently successful whenseeking antiderivatives of functions. Sometimes this is a simple problem, since it will beapparent that the function you wish to integrate is a derivative in some straightforwardway. For example, faced withintegraldisplayx10dxwe realize immediately that the derivative ofx11will supply anx10: (x11)′= 11x10. Wedon’t want the “11”, but constants are easy to alter, because differentiation “ignores” themin certain circumstances, soddx111x11=11111x10=x10.From our knowledge of derivatives, we can immediately write down a number of an-tiderivatives. Here is a list of those most often used:integraldisplayxndx=xn+1n+ 1+C,ifnnegationslash=−1integraldisplayx−1dx= ln|x|+Cintegraldisplayexdx=ex+Cintegraldisplaysinx dx=−cosx+C149
150Chapter 8 Techniques of Integrationintegraldisplaycosx dx= sinx+Cintegraldisplaysec2x dx= tanx+Cintegraldisplaysecxtanx dx= secx+Cintegraldisplay11 +x2dx= arctanx+Cintegraldisplay1√1−x2dx= arcsinx+CNeedless to say, most problems we encounter will not be so simple. Here’s a slightly morecomplicated example: findintegraldisplay2xcos(x2)dx.This is not a “simple” derivative, but a little thought reveals that it must have come froman application of the chain rule. Multiplied on the “outside” is 2x, which is the derivativeof the “inside” functionx2. Checking:ddxsin(x2) = cos(x2)ddxx2= 2xcos(x2),sointegraldisplay2xcos(x2)dx= sin(x2) +C.Even when the chain rule has “produced” a certain derivative, it is not always easy tosee. Consider this problem:integraldisplayx3radicalbig1−x2dx.There are two factors in this expression,x3andradicalbig1−x2, but it is not apparent that thechain rule is involved. Some clever rearrangement reveals that it is:integraldisplayx3radicalbig1−x2dx=integraldisplay(−2x)parenleftbigg−12parenrightbigg(1−(1−x2))radicalbig1−x2dx.This looks messy, but we do now have something that looks like the result of the chainrule: the function 1−x2has been substituted into−(1/2)(1−x)√x, and the derivative
8.1 Substitution151of 1−x2,−2x, multiplied on the outside. If we can find a functionF(x) whose derivativeis−(1/2)(1−x)√xwe’ll be done, since thenddxF(1−x2) =−2xF′(1−x2) = (−2x)parenleftbigg−12parenrightbigg(1−(1−x2))radicalbig1−x2=x3radicalbig1−x2But this isn’t hard:integraldisplay−12(1−x)√x dx=integraldisplay−12(x1/2−x3/2)dx(8.1)=−12parenleftbigg23x3/2−25x5/2parenrightbigg+C=parenleftbigg15x−13parenrightbiggx3/2+C.