121616949-math.170 - x 2 d dx x 2 = 2 x cos x 2 so Z 2 x...

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156 Chapter 8 Techniques of Integration Z sin x dx = - cos x + C Z cos x dx = sin x + C Z sec 2 x dx = tan x + C Z sec x tan x dx = sec x + C Z 1 1 + x 2 dx = arctan x + C Z 1 1 - x 2 dx = arcsin x + C Needless to say, most problems we encounter will not be so simple. Here’s a slightly more complicated example: find Z 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(
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Unformatted text preview: x 2 ) d dx x 2 = 2 x cos( x 2 ) , so Z 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: Z x 3 p 1-x 2 dx. There are two factors in this expression, x 3 and √ 1-x 2 , but it is not apparent that the chain rule is involved. Some clever rearrangement reveals that it is: Z x 3 p 1-x 2 dx = Z (-2 x ) ±-1 2 ¶ (1-(1-x 2 )) p 1-x 2 dx....
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