Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (h)

MATH105 April 2012

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Question 08 (h)
Short-answer questions! No credit will be given for the answer (even if it is correct) without the accompanying work. Evaluate the indefinite integral $\int {\frac {\sqrt {25x^{2}-4}}{x}}\,dx.$

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
You will need your trig identities ready for this one. Recall $\sec ^{2}\theta -1=\tan ^{2}\theta$.

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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. To evaluate $I=\int {\frac {\sqrt {25x^{2}-4}}{x}}dx$, we will use trigonometric substitution. The part with the square root is $25x^{2}-4$. We only have freedom over what we do to x, and we'd like to use a trig identity to help deal with the square root. We recall $\sec ^{2}\theta -1=\tan ^{2}\theta$. If $25x^{2}=4\sec ^{2}\theta$ then ${\sqrt {25x^{2}-4}}={\sqrt {4\tan ^{2}\theta }}=2\tan \theta$. Let's make the change of variables $x={\frac {2}{5}}\sec \theta$ (by solving $25x^{2}=4\sec ^{2}\theta$ for x. If $x={\frac {2}{5}}\sec \theta$ then $dx={\frac {2}{5}}\tan \theta \sec \theta d\theta$: $I=\int {\frac {\overbrace {2\tan \theta } ^{\sqrt {25x^{2}-4}}}{\underbrace {{\frac {2}{5}}\sec \theta } _{x}}}\underbrace {{\frac {2}{5}}\tan \theta \sec \theta d\theta } _{dx}=2\int \tan ^{2}\theta d\theta$. Now, we can use a trig identity: $I=2\int \tan ^{2}\theta d\theta =2\int (\sec ^{2}\theta -1)d\theta =2(\tan \theta -\theta )+C$. Finally, we need to go back in terms of x. If $x={\frac {2}{5}}\sec \theta$ then $\sec \theta ={\frac {5}{2}}x$ and $1+\tan ^{2}\theta =\sec ^{2}\theta ={\frac {25}{4}}x^{2}$. This means we can replace $\tan \theta$ by ${\sqrt {{\frac {25}{4}}x^{2}-1}}$. Also, $\theta =\operatorname {arcsec}({\frac {5}{2}}x)$. Thus, our final answer is: $I=2({\sqrt {{\frac {25}{4}}x^{2}-1}}-\operatorname {arcsec}({\frac {5}{2}}x))+C$.

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