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Unformatted text preview: 8 Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with integraldisplay x 10 dx we realize immediately that the derivative of x 11 will supply an x 10 : ( x 11 ) ′ = 11 x 10 . We don’t want the “11”, but constants are easy to alter, because differentiation “ignores” them in certain circumstances, so d dx 1 11 x 11 = 1 11 11 x 10 = x 10 . From our knowledge of derivatives, we can immediately write down a number of an tiderivatives. Here is a list of those most often used: integraldisplay x n dx = x n +1 n + 1 + C, if n negationslash = − 1 integraldisplay x − 1 dx = ln  x  + C integraldisplay e x dx = e x + C integraldisplay sin x dx = − cos x + C 149 150 Chapter 8 Techniques of Integration integraldisplay cos x dx = sin x + C integraldisplay sec 2 x dx = tan x + C integraldisplay sec x tan x dx = sec x + C integraldisplay 1 1 + x 2 dx = arctan x + C integraldisplay 1 √ 1 − x 2 dx = arcsin x + C 8Ùb ×Ø iØÙØ iÓÒ Needless to say, most problems we encounter will not be so simple. Here’s a slightly more complicated example: find integraldisplay 2 x cos( x 2 ) dx. This is not a “simple” derivative, but a little thought reveals that it must have come from an application of the chain rule. Multiplied on the “outside” is 2 x , which is the derivative of the “inside” function x 2 . Checking: d dx sin( x 2 ) = cos( x 2 ) d dx x 2 = 2 x cos( x 2 ) , so integraldisplay 2 x cos( x 2 ) dx = sin( x 2 ) + C. Even when the chain rule has “produced” a certain derivative, it is not always easy to see. Consider this problem: integraldisplay x 3 radicalbig 1 − x 2 dx. There are two factors in this expression, x 3 and radicalbig 1 − x 2 , but it is not apparent that the chain rule is involved. Some clever rearrangement reveals that it is: integraldisplay x 3 radicalbig 1 − x 2 dx = integraldisplay ( − 2 x ) parenleftbigg − 1 2 parenrightbigg (1 − (1 − x 2 )) radicalbig 1 − x 2 dx. This looks messy, but we do now have something that looks like the result of the chain rule: the function 1 − x 2 has been substituted into − (1 / 2)(1 − x ) √ x , and the derivative 8.1 Substitution 151 of 1 − x 2 , − 2 x , multiplied on the outside. If we can find a function F ( x ) whose derivative is − (1 / 2)(1 − x ) √ x we’ll be done, since then d dx F (1 − x 2 ) = − 2 xF ′ (1 − x 2 ) = ( − 2 x ) parenleftbigg − 1 2 parenrightbigg (1 − (1 − x 2 )) radicalbig 1 − x 2 = x 3 radicalbig 1 − x 2 But this isn’t hard: integraldisplay − 1 2 (1 − x ) √ x dx = integraldisplay − 1 2 ( x 1 / 2 − x 3 / 2 ) dx (8 . 1) = − 1 2 parenleftbigg 2 3 x 3 / 2 − 2 5 x 5 / 2 parenrightbigg + C = parenleftbigg 1 5 x − 1 3 parenrightbigg...
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This note was uploaded on 12/01/2011 for the course MATH 305 taught by Professor Guichard during the Fall '11 term at Whitman.
 Fall '11
 GUICHARD

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