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Unformatted text preview: 8.2 INTEGRATION BY PARTS We begin with the formula for the derivative of a product f ( x ) g ( x ) + f ( x ) g ( x ) = ( f g ) ( x ). Integrating both sides, we get f ( x ) g ( x ) dx + f ( x ) g ( x ) dx = ( f g ) ( x ) dx . Since ( f g ) ( x ) dx = f ( x ) g ( x ) + C , we have f ( x ) g ( x ) dx + f ( x ) g ( x ) dx = f ( x ) g ( x ) + C , and therefore f ( x ) g ( x ) dx = f ( x ) g ( x ) f ( x ) g ( x ) dx + C . Since the calculation of f ( x ) g ( x ) dx will yield its own arbitrary constant, there is no reason to keep the constant C . We therefore drop it and write (8.2.1) f ( x ) g ( x ) dx = f ( x ) g ( x ) f ( x ) g ( x ) dx . 8.2 INTEGRATION BY PARTS 451 This formula, called the formula for integration by parts , enables us to find f ( x ) g ( x ) dx by finding f ( x ) g ( x ) dx instead. Of course, it is of practical use only if the second integral is easier to calculate than the first. In practice we usually set u = f ( x ), dv = g ( x ) dx . Then du = f ( x ) dx , v = g ( x ). Now, with these substitutions, the formula for integration by parts can be written (8.2.2) u dv = uv v du . Success with this formula depends on choosing u and dv so that v du is easier to calculate than u dv . Example 1 Calculate x e x dx . SOLUTION We want to separate x from e x . Setting u = x , dv = e x dx , we have du = dx , v = e x . Accordingly, x e x dx = u dv = uv v du = xe x e x dx = x e x e x + C . Our choice of u and dv worked out very well. Does the choice of u and dv make a difference? Suppose we had set u = e x , dv = x dx ; then we would have had du = e x dx , v = 1 2 x 2 . 452 CHAPTER 8 TECHNIQUES OF INTEGRATION Integration by parts would then have given xe x dx = u dv = uv v du = 1 2 x 2 e x 1 2 x 2 e x dx , an integral more complicated than the one we started with. Clearly, the proper selection of u and dv is important. Example 2 Calculate x sin 2 x dx . SOLUTION Setting u = x , dv = sin 2 x dx , we have du = dx , v = 1 2 cos 2 x . Therefore, x sin 2 x dx = 1 2 x cos 2 x 1 2 cos 2 x dx = 1 2 x cos 2 x + 1 4 sin 2 x + C . You can verify that if we had set u = sin 2 x , dv = x dx , then we would run into the same kind of difficulty that we saw in Example 1. In Examples 1 and 2 there was only one effective choice for u and dv . However, in some cases there may be more than one way to choose u and dv . Example 3 Calculate x ln x dx . SOLUTION Setting u = ln x , dv = x dx , we have du = 1 x dx , v = x 2 2 . Then x ln x dx = u dv = uv v du = x 2 2 ln x 1 x x 2 2 dx = 1 2 x 2 ln x 1 2 x dx = 1 2 x 2 ln x 1 4 x 2 + C . ALTERNATE SOLUTION This time we set u = x ln x , dv = dx , so that du = (1 + ln x ) dx , v = x ....
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston  Downtown.
 Spring '10
 SMITH

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