Science:Math Exam Resources/Courses/MATH101/April 2005/Question 03 (c)

MATH101 April 2005

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Question 03 (c)
Full-Solution Problem. Justify your answer and show all your work. Simplification of your answer is not required. Evaluate the following integral $\displaystyle \int {\frac {dx}{(5-4x-x^{2})^{3/2}}}$

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?

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.

Hint 1
Try to modify the integrand in order to perform a trigonometric substitution.

Hint 2
Complete the square in the denominator and make an appropriate trigonometric substitution.

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.

Solution

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 want to integrate: $\int {\frac {dx}{(5-4x-x^{2})^{3/2}}}.$ The first step is to complete the square in the denominator: ${\begin{aligned}\int {\frac {dx}{(9-4-4x-x^{2})^{3/2}}}&=\int {\frac {dx}{(9-(x+2)^{2})^{3/2}}}\\&=\int {\frac {1}{27}}\;{\frac {dx}{(1-{\frac {(x+2)^{2}}{9}})^{3/2}}}\\&={\frac {1}{27}}\int {\frac {dx}{(1-({\frac {x+2}{3}})^{2})^{3/2}}}\end{aligned}}$ Now we substitute: $y={\frac {x+2}{3}}$ $dy={\frac {dx}{3}}$ $\displaystyle 3dy=dx,$ So the integral simplifies to ${\frac {3}{27}}\int {\frac {dy}{(1-y^{2})^{3/2}}}={\frac {1}{9}}\int {\frac {dy}{(1-y^{2})^{3/2}}}$ We now perform a trigonometric substitution (this step is only valid for |y| smaller than 1, which we know is true because 1-y^2 must be positive): $\displaystyle y=\sin(t)$ $\displaystyle dy=\cos(t)dt$ So the above integral becomes: ${\begin{aligned}{\frac {1}{9}}\int {\frac {\cos(t)dt}{(1-\sin ^{2}(t))^{3/2}}}&={\frac {1}{9}}\int {\frac {\cos(t)dt}{(\cos ^{2}(t))^{3/2})}}\\&={\frac {1}{9}}\int {\frac {dt}{\cos ^{2}(t)}}\\&={\frac {1}{9}}\int \sec ^{2}(t)dt\\&={\frac {1}{9}}\tan(t)+C\\&={\frac {1}{9}}\tan(\arcsin(y))+C\\&={\frac {1}{9}}\tan(\arcsin({\frac {x+2}{3}}))+C\end{aligned}}$ But for any u, we have $\tan(\arcsin(u))={\frac {u}{\sqrt {1-u^{2}}}},$ So the above quantity is equal to $={\frac {1}{9}}\;{\frac {\frac {x+2}{3}}{\sqrt {1-({\frac {x+2}{3}})^{2}}}}$ $={\frac {1}{9}}\;{\frac {x+2}{\sqrt {5-4x-x^{2}}}}.$ So we have $\int {\frac {dx}{(5-4x-x^{2})^{3/2}}}={\frac {1}{9}}\;{\frac {x+2}{\sqrt {5-4x-x^{2}}}}.$ This is not required in a test, but we can check our answer by differentiating: ${\frac {d}{dx}}\;{\frac {1}{9}}\;{\frac {x+2}{\sqrt {5-4x-x^{2}}}}$ ${\begin{aligned}&={\frac {1}{9}}\;{\frac {{\sqrt {5-4x-x^{2}}}-{\frac {1}{2}}(x+2)(5-4x-x^{2})^{-1/2}(-2x-4)}{(5-4x-x^{2})}}\\&={\frac {1}{9}}\;{\frac {5-4x-x^{2}+x^{2}+2x+2x+4}{(5-4x-x^{2})^{3/2}}}\\&={\frac {1}{9}}\;{\frac {9}{(5-4x-x^{2})^{3/2}}}\\&={\frac {1}{(5-4x-x^{2})^{3/2}}}\end{aligned}}$

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