Science:Math Exam Resources/Courses/MATH105/April 2017/Question 02 (b)

MATH105 April 2017

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q3  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q5 (e)  •  Q6 (a)  •  Q6 (b)  •  

   Other MATH105 Exams

  •  April 2009  •  April 2011  •  April 2010  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2017  •  April 2018  •  

Question 02 (b)
Evaluate $\int {\frac {(x-4)^{2}}{(9+8x-x^{2})^{\frac {3}{2}}}}dx$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
By looking at the form of this integral, you should immediately suspect that the integral should be solved by trigonometric substitution. The first step in applying the method of trigonometric substitution is to complete the square in the denominator. Try rewriting the denominator as $(25-(x-4)^{2})^{3/2}$. Make sure you are able to complete the square on your own.

Checking a solution serves two purposes: helping you if, after having used the hint, 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. The presence of the square root of the quadratic in the denominator suggests a trigonometric substitution. However, some work is needed in order to set up the trigonometric substitution. The first step in applying the method of trigonometric substitution is to complete the square in the quadratic in the denominator. We have $-(x^{2}-8x)$ as the linear and quadratic terms, and to complete the square we would like a term of the form $-(x^{2}-8x+16)=-(x-4)^{2}$. Therefore, we rewrite our integral: $\int {\frac {(x-4)^{2}}{(9+8x-x^{2})^{3/2}}}\,dx=\int {\frac {(x-4)^{2}}{(25-16+8x-x^{2})^{3/2}}}\,dx$ which can be rewritten as $\int {\frac {(x-4)^{2}}{(25-(x-4)^{2})^{3/2}}}$ Luckily, the term x-4 appears both in the numerator and denominator, which will simplify the calculations somewhat. We make the substitution u = x-4: $\int {\frac {u^{2}}{(25-u^{2})^{3/2}}}$ We are finally ready to make a trigonometric substitution. Because the term in the square root in the denominator is $25-u^{2}$, we make the substitution $u=5\sin \theta$. Thus $du=5\cos \theta d\theta$. Our integral becomes $\int {\frac {125\sin ^{2}\theta \cos \theta }{(25-25\sin ^{2}\theta )^{3/2}}}\,d\theta .$ Now, we notice that $25-25\sin ^{2}\theta$ is $25\cos ^{2}\theta$. Raising this to the power of 3/2 gives $125\cos ^{3}\theta$. So the integral simplifies to $\int {\frac {125\sin ^{2}\theta \cos \theta }{125\cos ^{3}\theta }}\,d\theta .$ We cancel out $125\cos \theta$ from the numerator and denominator to get $\int {\frac {\sin ^{2}\theta }{\cos ^{2}\theta }}\,d\theta$ We now recognize the integrand as $\tan ^{2}\theta$. The next step is to apply the trigonometric identity $\tan ^{2}\theta =\sec ^{2}\theta -1$. Make sure you have this identity memorized. The integral becomes $\int \sec ^{2}\theta -1\,d\theta$ which we recognize as $\tan \theta -\theta +C.$ Now, we need to rewrite this in terms of u, and then in terms of x. Recall that $u=5\sin \theta$ so $\theta =\arcsin {\frac {u}{5}}$. Thus $\tan \theta =\tan \arcsin {\frac {u}{5}}$. To compute this, draw a right triangle with angle $\theta$, opposite side u, and hypotenuse 5. The adjacent side will have length ${\sqrt {25-u^{2}}}$ and therefore $\tan \theta ={\frac {u}{\sqrt {25-u^{2}}}}+C$. Thus our integral is ${\frac {u}{\sqrt {25-u^{2}}}}-\arcsin {\frac {u}{5}}$. Finally, we remember that u = x-4, so we end up with the integral ${\frac {x-4}{\sqrt {25-(x-4)^{2}}}}-\arcsin({\frac {x-4}{5}})+C.$ Answer: $\color {blue}{\frac {x-4}{\sqrt {25-(x-4)^{2}}}}-\arcsin \left({\frac {x-4}{5}}\right)+C$