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# E xample 7 calculate s olution by inspection or

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Unformatted text preview: ring the coefﬁcients of ￿￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿, and get a system of equations. Alternatively, we may clear the denominators, then use particular values for ￿ to create systems of equations; this method is extremely useful when using roots of ￿￿￿￿ to eliminate terms. 6 WATERLOO SOS E XAM -AID: MATH138 M IDTERM Step 4: Integrate term by term from step 3. Recall that ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿￿ ￿ ￿ if ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿ ￿ ￿ if ￿ ￿ ￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿ For the quadratic case, we ﬁrst complete the square. Once we are in the form ￿ ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ we can break up the integral as ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿￿ The ﬁrst integral is simple, since differentiating ￿￿ ￿ ￿ ￿ gives a ￿￿ term, so ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿ ￿ ￿￿ ￿￿ The second integral can be completed using a trigonometric substitution ￿ ￿ ￿ ￿￿￿ ￿, with ￿￿ ￿ ￿ ￿￿￿￿ ￿: ￿ ￿ ￿ ￿ ￿ ￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿ ￿￿￿￿￿ ￿￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿ which we can solve using previous techniques. Notice that in the case ￿ ￿ ￿, this reduces to ￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿￿￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿ as expected. Rationalizing Substitution: If the integrand contains a term of the form 3 Strategy for Integration (7.5) ￿ ￿ ￿ ￿ ￿￿￿, then the substitution ￿ ￿ ￿ ￿ ￿￿￿ may be effective. The textbook uses a four-step strategy to solve integrals, which is very useful, but not perfect - integration is hard! (1) Simplify the integrand if possible. (2) Look for an obvious substitution or antiderivative. (3) Attempt to use one of the ﬁve methods you have learned: (i) Integration by parts (ii) Trigonometric integration (iii) Trigonometric substitution (iv) Partial fractions (v) Rationalizing substitution 7 WATERLOO SOS E XAM -AID: MATH138 M IDTERM (4) Try again. E XAMPLE 1. Calculate ￿ ￿￿ ￿￿￿ ￿￿ ￿ ￿￿ ￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿ S OLUTION. The strategy is not immediately obvious. There is no clear simpliﬁcation to be made, no trigonometric functions to substitute, no radicals, and no rational functions. However, notice that the derivative of ￿ ￿￿￿￿￿￿ ￿ will yield a ￿￿￿￿ term, so we try for a direct antiderivative. To get ￿￿￿￿￿￿￿ ￿ ￿ ￿￿ in the denominator, we use a logarithm: ￿￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿￿ Taking the derivative, we see by chain rule that ￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿ ￿￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ Thus, to get to our integrand, we require a negative sign: ￿ ￿￿ ￿￿ ￿ ￿ ￿￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿ ￿ ￿￿ ￿￿￿￿￿￿￿￿ ￿ ￿ ￿￿ E XAMPLE 2. Calculate ￿ ￿￿￿ ￿￿￿ ￿￿ ￿￿￿ S OLUTION. We have a product of functions, so we try integration by parts: ￿ ￿ ￿￿￿ ￿￿ ￿ ￿￿￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿￿￿ ￿￿ ￿ ￿￿ ￿ ￿￿￿￿ ￿￿ Then ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿ ￿￿ ￿ ￿￿￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿ ￿￿￿ ￿￿ ￿￿ The integral is no easier than what we started with, but we can use integration by parts a second time, to get a recurring integral: ￿ ￿ ￿￿￿ ￿￿ ￿ ￿￿￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿ ￿￿ ￿ ￿￿ ￿ ￿￿￿￿ ￿￿ Then ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿￿ ￿￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿ 8 WATERLOO SOS E XAM -AID: MATH138 M IDTERM We solve for our original integral using simple algebra now. ￿ ￿ ￿ ￿...
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## This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.

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