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Unformatted text preview: ￿ ￿￿ ￿￿￿ ￿ S OLUTION. The strategy is not immediately obvious. There is no clear simpliﬁcation to be made, no trigonometric functions, no radicals, and no rational functions. We might be able to use integration by parts, using 13 WATERLOO SOS E XAM -AID: MATH138 M IDTERM the “1” trick. ￿ ￿ ￿￿￿￿￿ ￿ ￿￿ ￿￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿ ￿ ￿￿ ￿￿￿ Then ￿ ￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ Notice ￿￿ ￿￿ is a rational function, so we can try using integration by parts. By our algorithm, we must ﬁrst apply polynomial division since the degree of the numerator is at least the degree of the denominator. We get ￿￿ ￿ ￿￿￿ ￿ ￿ ￿￿￿ ￿ ￿ ￿￿ Notice the denominator is already factored into 1 term, so we have the partial fraction decomposition. ￿ ￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿ ￿￿￿￿ ￿￿ ￿ ￿￿￿￿￿￿￿￿ ￿￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿ ￿ ￿ ￿ ￿ ￿ E XERCISE 2. Calculate ￿ ￿ ￿ ￿￿￿ ￿￿￿ ￿￿￿ ￿￿ S OLUTION. This is a simple integration by parts: ￿ ￿￿ ￿￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿￿ ￿￿ ￿￿ ￿ so that ￿ ￿ ￿ Using another integration by parts, ￿￿ ￿￿ ￿￿￿ ￿￿￿ ￿￿￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ 14 WATERLOO SOS E XAM -AID: MATH138 M IDTERM so that ￿ ￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿￿ ￿ ￿￿￿ ￿￿￿ ￿￿￿ ￿￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿￿￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿ ￿￿￿ ￿￿ E XERCISE 3. Calculate ￿ ￿ ￿￿ ￿￿ ￿ ￿ ￿￿ in two ways, by making the substitution ￿ ￿ ￿ ￿ ￿￿ and by a suitable trigonometric substitution. S OLUTION. Making the substitution ￿ ￿ ￿ ￿ ￿￿ , we get ￿￿ ￿ ￿￿￿ ￿￿ and ￿￿ ￿ ￿ ￿ ￿￿ . Then ￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿￿￿￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ Alternatively, we could make the trigonometric substitution ￿ ￿ ￿￿￿ ￿ so that ￿￿ ￿ ￿￿￿ ￿ ￿￿. Then ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿￿￿￿ ￿ ￿￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿￿￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿￿ ￿ ￿￿ ￿￿￿ ￿￿￿ ￿ ￿￿￿￿ ￿￿ ￿￿￿ Now we make the substitution ￿ ￿ ￿￿￿ ￿ so that ￿￿ ￿ ￿ ￿￿￿ ￿ ￿￿. ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿￿ ￿ ￿￿￿ ￿ ￿ ￿￿￿￿ ￿ ￿ ￿￿￿ ￿ ￿ ￿ ￿ ￿￿￿￿￿ ￿￿￿ ￿ ￿￿ ￿ ￿ To return to our original variable, we must construct the triangle generated by ￿￿￿ ￿ ￿ ￿ , yielding ￿￿￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ . Then ￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ As expected, our answers coincide. 15 WATERLOO SOS E XAM -AID: MATH138 M IDTERM E XERC...
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