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As always with substitution emember to change the

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Unformatted text preview: ￿￿￿ ￿ ￿￿ to pick up the extra factor of ￿￿￿ ￿, and we are left with a polynomial. (iii) If both the powers of sine and cosine are even, we could try using the half-angle identites ￿￿￿￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿￿￿￿ ￿ ￿￿￿￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿￿￿￿ ￿ For combinations of powers of ￿￿￿ ￿ and ￿￿￿ ￿: (i) If there is at least one term of secant and the power of tangent 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 factors of ￿￿￿ ￿ ￿￿￿ ￿, and we are left with a polynomial. (ii) Similarly, if there is at least two terms of secant and the power of secant is even, say we have a term of the form ￿￿￿￿￿ ￿ with ￿ ￿ ￿, 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. (iii) If there is an even power of tangent and an odd power of secant, you can use ￿￿￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿ ￿ and express the integrand in terms of secant, then use integration by parts. (iv) In other cases, there is no general algorithm, but we might be able to use identities, integration by parts, and the following: ￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿￿ ￿￿ ￿ ￿￿ ￿￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿￿ ￿ ￿ ￿￿￿ ￿￿ ￿ ￿￿ For combinations of ￿￿￿ ￿￿ and ￿￿￿ ￿￿: 4 WATERLOO SOS E XAM -AID: MATH138 M IDTERM (i) Given ￿￿￿ ￿￿ ￿￿￿ ￿￿, use ￿￿￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿￿￿ ￿ (ii) Given ￿￿￿ ￿￿ ￿￿￿ ￿￿, use ￿￿￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿￿￿ ￿ (iii) Given ￿￿￿ ￿￿ ￿￿￿ ￿￿, use ￿￿￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿￿￿ ￿ If you are using one of the substitution techniques, remember to change the bounds of integration. If it is an indefinite integral, remember to return to the original variable. Trigonometric Substitution: ￿ When the integrand has a term of the form ￿￿￿ ￿ ￿￿ , we may be tempted to use a substitution like ￿ ￿ ￿￿￿ ￿ ￿￿ . However, the change of variables theorem will yield a term ￿￿￿ ￿￿, which may be difficult to handle. Instead, we will make use of some trigonometric identities, employing what is known as an inverse substitution, provided we restrict our domain to ensure we have a one-to-one function. ￿ (i) Given ￿￿ ￿ ￿￿ , use ￿ ￿ ￿ ￿￿￿ ￿ for ￿ ￿ ￿ ￿ ￿ and the identity ￿ ￿ ￿￿￿￿ ￿ ￿ ￿￿￿￿ ￿. Then ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿, and ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿￿￿￿ ￿ ￿ ￿ ￿￿￿ ￿￿ ￿ (ii) Given ￿￿ ￿ ￿￿ , use ￿ ￿ ￿ ￿￿￿ ￿ for ￿ ￿ ￿ ￿ ￿ ￿ and the identity ￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿ ￿. Then ￿ ￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿, and ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿ Alternatively, you can use ￿ ￿ ￿ ￿￿￿￿ ￿ and the identity ￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿. Then ￿￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿￿, and ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿￿￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿￿ ￿ ￿ (iii) Given ￿￿ ￿ ￿￿ , use ￿ ￿ ￿ ￿￿￿ ￿ for ￿ ￿ ￿ ￿ ￿ or ￿ ￿ ...
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This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.

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