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Published byPosy Sharp Modified over 9 years ago
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Continuing with Integrals of the Form: & Partial Fractions Chapter 7.3 & 7.4
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We use inverse substitution with x = g(t), and dx = g’(t)dt to transform:
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Evaluate: Chapter 7.3 & 7.4 March 29, 2007
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Back to x?:
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Suppose you used the substitution: Complete the problem if the integration resulted in: How would your solution change if the substitution was instead?
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Determine the integral obtained when using the appropriate trig. change of variable:
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7.4 Integrating Rational Functions or “Undoing” Addition! First check is to be sure the Rational Function is in reduced form. In reduced form, the degree of the numerator is less than the degree of the denominator. If not, the first step is to simplify the expression using LONG division…….. So for example, the rational function in the integral below is NOT in reduced form.
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Example:
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The Arctangent formula (also see day 9 notes) The “new” formula: When to use? If the polynomial in the denominator does not have real roots (b 2 -4ac < 0) then the integral is an arctangent, we complete the square and integrate…. For example:
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What if the polynomial has real roots? That means we can factor the polynomial and “undo” the addition! To add fractions we find a common denominator and add: we’ll work the other way…. The denominator of our rational function factors into (t - 4)(t +1) So in our original “addition,” the fractions were of the form:
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Integrating with the Partial Fraction Decomposition:
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Examples:
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