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Unit3 - Unit 3 Integration Techniques We have learned how...

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Unit 3: Integration Techniques We have learned how to integrate functions of the following friendly forms, and have learned how to use the method of substitution to transform some problems into these forms: integraltext u n du = u n +1 n +1 + C, n negationslash = - 1 integraltext 1 u du = ln | u | + C integraltext e u du = e u + C But sometimes this is not enough. For instance, if we wish to evaluate something like integraltext x ln( x ) dx , nothing we have learned so far will work. For something like this we turn to a new method, namely that of integration by parts , which is sometimes called “double substitution”. The derivation of the formula is easily seen to be a consequence of the product rule. When u and v are functions of x , we have d dx [ uv ] = uv prime + vu prime where, of course, v prime = dv dx and u prime = du dx . Antidifferentiating both sides of this equation with respect to x we get uv = integraldisplay uv prime dx + integraldisplay vu prime dx which can be rearranged to integraldisplay uv prime dx = uv - integraldisplay vu prime dx This is the formula we use. Notice that since v prime = dv dx , then v prime dx = dv dx dx = dv and since u prime = du dx , then u prime dx = du dx dx = du . We get: 1
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Definition 3.1. The integration by parts formula: If u and v are differentiable functions then, integraltext udv = uv - integraltext vdu This formula is useful when integraltext vdu is an integral which is easier to evaluate than the one we started with, integraltext udv . So when facing a difficult integral, we consider whether we can recognize the form integraltext udv for some u and v . The following steps may make it a bit easier to use this method. Step 1: Write the given integral in the form integraltext f ( x ) g ( x ) dx so that you can identify the two functions f ( x ) and g ( x ). Note that if there appears to be only one function in the integrand, then you may need to identify the second one to be the constant function, g ( x ) = 1. Step 2: Introduce the functions u ( x ) and v ( x ) by letting: u = f ( x ) and dv = g ( x ) dx Now, to use the integration by parts formula, we need to find du and v . That is, we differentiate u to get: du dx = f prime ( x ) du = f prime ( x ) dx and we integrate dv to get: v = integraltext dv v = integraltext g ( x ) dx Step 3: Use the formula to re-express the original integral: replace integraltext udv with uv - integraltext vdu Step 4: Evaluate the new expression (by finding integraltext vdu ). Note: When we find v = integraltext dv = integraltext v prime dx in Step 2, we don’t need to include an arbitrary constant. This is because when we use v in our formula, we will still have an indefinite integral appearing in the expression. All arbitrary constants will be collected together as one when we perform the last integra- tion. 2
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Of course, the main difficulty in this method is determining the choice of u and dv . In Step 1, we recognize the integrand as the product of 2 terms.
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