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Calculate s olution we have a product of functions

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Unformatted text preview: ￿ ￿￿ and the identity ￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿ ￿. ￿ Then ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿ ￿ ￿￿, and ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿ Alternatively, you can use ￿ ￿ ￿ ￿￿￿￿ ￿ and the identity ￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿. Then ￿￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿￿, and ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿￿ Note: Although we have very specific bounds, their choice is not very important: one could choose ￿￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ for the restriction when dealing with ￿￿￿ ￿, for example. We need only ensure that the func￿ ￿ tion is one-to-one. Furthermore, we could have used ￿￿￿ ￿, ￿￿￿ ￿, and ￿￿￿ ￿ instead of ￿￿￿ ￿, ￿￿￿ ￿, and ￿￿￿ ￿ respectively, since they have the same desired properties, but they would work on a different interval. For example, ￿￿￿ ￿ is not one-to-one for ￿ ￿ ￿ ￿ ￿ ￿ , but is one-to-one for ￿ ￿ ￿ ￿ ￿ . ￿ ￿ As always with substitution, emember to change the bounds of integration. If it is an indefinite integral, remember to return to the original variable. However, be careful - since we used an inverse substitution, the old variable is a function of the new one rather than the other way around. When dealing with trigonometric substitution, we often end up with a trigonometric term such as ￿￿￿ ￿ when we began with a substitution like ￿ ￿ ￿ ￿￿￿ ￿. Rather than use ￿￿￿￿￿￿￿￿￿￿￿ ￿ ￿￿, you will be required to simplify the expression into some￿ thing that requires no trigonometric functions. We do this by using the information garnered from the right-triangle produced from our substitution equation. 5 WATERLOO SOS E XAM -AID: MATH138 M IDTERM Partial Fractions: When confronted with a rational function (a ratio of polynomials), we can try using partial fractions in a somewhat algorithmic way - as long as we can factor polynomials, divide polynomials, and solve systems ￿ ￿ of equations. Suppose you must find ￿ ￿￿￿ ￿￿, for ￿ ￿￿￿ ￿ ￿￿￿￿ , where ￿ ￿￿￿ and ￿￿￿￿ are polynomials in ￿￿ ￿ ￿. We will apply the following algorithm to express the solution as a sum of rational functions, lograrithms of polynomials, and arctan of polynomials. Step 1: If necessary, use polynomial division to express ￿ ￿￿￿ ￿ ￿ ￿￿￿ are polynomials with ￿￿￿ ￿ ￿ ￿￿￿ ￿. ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ , where ￿ ￿￿￿ and Step 2: Completely factor the denominator ￿￿￿￿. A theorem from algebra guarantees a factorization into linear terms (of the form ￿￿ ￿ ￿) and irreducible quadratic terms (of the form ￿￿￿ ￿ ￿￿ ￿ ￿ such that ￿￿ ￿ ￿￿￿ ￿ ￿). Common tricks to use here are common factors, difference of squares, sum and difference of cubes, quadratic formula, and temporarily considering ￿￿ as the variable instead of ￿. Step 3: Express ￿ ￿￿ ￿ ￿￿￿ ￿ as a sum of partial fractions of the form ￿ ￿￿￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿￿ or ￿￿￿￿ where ￿ ￿ ￿ ￿ ￿ and ￿￿￿ ￿ ￿￿￿ appears as a factor of ￿￿￿￿, or ￿ ￿ ￿ ￿ ￿ and ￿￿￿￿ ￿ ￿￿ ￿ ￿￿￿ appears as a factor of ￿￿￿￿. To be explicit: (i) If ￿￿￿￿ has a linear term with multiplicity ￿, say ￿￿￿ ￿ ￿￿￿ (and ￿￿￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿￿￿￿), then add ￿￿ ￿￿ ￿￿ ￿ ￿ ￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿ ￿￿￿ ￿ ￿￿￿ (ii) If ￿￿￿￿ has a quadratic term with multiplicity ￿, say ￿￿￿￿ ￿ ￿￿ ￿ ￿￿￿ (and ￿￿￿￿ ￿ ￿￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿￿￿￿), then add ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿￿ ￿￿￿ ￿￿￿￿ ￿ ￿￿ ￿ ￿￿￿ We can solve for the coefficients by clearing the denominators using cross-multiplication, then compa...
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This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.

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