MATH101 April 2008
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Question 03 (c)
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Full-Solution Problem. Justify your answers and show all your work. Simplification of answers is not required.
Evaluate the following integral.
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Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.
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Hint 1
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There are (at least) two distinct ways to solve this.
Option 1: Start with a trigonometric substitution.
Option 2: Use partial fractions.
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Hint 2
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Option 1: Your first trigonometric substitution should make use of the fact that
Option 2: How do set up partial fraction when the denominator does not factor into linear factors?
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Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
- If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
- If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
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Solution
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Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies.
We recognize a denominator of the form and so we attempt a trigonometric substitution of the form
Then the integral becomes
To solve this integral we use another substitution:
to obtain
To revert back to an expression with we have to calculate
So we use a reference triangle. The sides and are given from the substitution With Pythagoras we can then calculate the length of the remaining side of the triangle, which is . Hence our final answer is
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MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Partial fractions, MER Tag Trigonometric substitution, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag
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