Midterm-Exam Aid

Midterm-Exam Aid

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Unformatted text preview: easier to manipulate the integral using a change of variable. T HEOREM 1.2 [S UBSTITUTION R ULE ]. If ￿ ￿ is continuous on ￿￿￿ ￿￿ and ￿ is continuous on the range of ￿ ￿ ￿ ￿￿￿, then ￿ ￿ ￿ ￿ ￿￿ ￿￿￿￿￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿ ￿￿￿ ￿ ￿￿￿ 2 WATERLOO SOS E XAM -AID: MATH138 M IDTERM The important thing to note here is that we must change the bounds of integration, and we need to consider the term ￿ ￿ ￿￿￿. If we are solving an indefinite integral, we must change back to the original variables. If we are solving a definite integral, we have two options: (i) Solve the problem in the new variable, change back to the original variable, and evaluate at the original bounds; or (ii) Solve the problem in the new variable, change the bounds of integration, and evaluate at these new bounds. 2 Techniques for Integration (7.1 - 7.4) We will discuss the all various methods of integration, before looking at examples in the next section. Typically, you will not be told what method of integration to use on the exam, so you must get used to identifying integrals. Integration by Parts: Recall the product rule for differentiation: if ￿ and ￿ are differentiable functions, then ￿ ￿ ￿ ￿￿ ￿￿￿￿ ￿￿￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿￿ ￿￿ ￿￿ ￿￿ Taking integrals, using the Fundamental Theorem, we get ￿ ￿ ￿ ￿￿￿￿ ￿￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿￿ ￿￿￿ Rearranging, ￿ ￿ ￿￿￿￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿￿￿ ￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿￿ ￿￿￿ Another form you may be familiar with is obtained when we use the notation ￿ ￿ ￿ ￿￿￿ and ￿ ￿ ￿ ￿￿￿, to get ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿￿ Remember the bounds if you are dealing with a definite integral: ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿ ￿ Integration by parts typically works well when you are confronted with a product of functions, but sometimes can be applied when there is no clear product of functions, by taking ￿￿ ￿ ￿￿. This “1” trick is useful when the integrand can differentiate to a nice form when multiplied by ￿, for example ￿￿ ￿, since ￿ ￿ ￿￿ ￿￿￿ ￿￿￿ ￿ ￿ ￿ ￿ ￿. Another useful trick is to use integration by parts twice to end up with a formula involving our original integral, then using simple algebra to find the solution. This “recurring integral” trick is useful when the integrand involves a product of functions which are cylic in their derivatives/antderivatives, for example ￿￿ ￿￿￿ ￿, since taking the second derivative/integral of ￿￿ is ￿￿ and taking the second derivative/integral of ￿￿￿ ￿ is ￿ ￿￿￿ ￿. Trigonometric Integration: 3 WATERLOO SOS E XAM -AID: MATH138 M IDTERM When you have an integral involving certain combinations of trigonometric functions, it can be useful to use a substitution such as ￿ ￿ ￿￿￿ ￿ or ￿ ￿ ￿￿￿ ￿ to attempt to transform the integrand into a rational polynomial. We do this by using trigonometric identities. For combinations of powers of ￿￿￿ ￿ and ￿￿￿ ￿: (i) If the power of sine is odd, say we have a term of the form ￿￿￿￿￿￿ ￿, then we use the identity ￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿ ￿ to get ￿￿￿￿￿￿￿ ￿ ￿ ￿￿￿￿￿ ￿￿￿ ￿￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿￿ ￿￿￿ ￿￿￿ ￿￿ Making the substitution ￿ ￿ ￿￿￿ ￿, we get ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿ to pick up the extra factor of ￿￿￿ ￿, and we are left with a polynomial. (ii) Similarly, if the power of cosine is odd, say we have a term of the form ￿￿￿￿￿￿￿ ￿, then we use the identity ￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿ ￿ to get ￿￿￿￿￿￿￿ ￿ ￿ ￿￿￿￿￿ ￿￿￿ ￿￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿￿ ￿￿￿ ￿￿￿ ￿￿ Making the substitution ￿ ￿ ￿￿￿ ￿, we get ￿￿ ￿...
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This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.

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