Science:Math Exam Resources/Courses/MATH101/April 2016/Question 03 (b)

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MATH101 April 2016

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Question 03 (b)
Evaluate $\displaystyle \int {\frac {dx}{x^{2}{\sqrt {x^{2}-9}}}}$ . Your answer cannot contain any inverse trigonometric functions.

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
Try substitution $x=3\sec u$ .

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Solution
Substituting ${\color {OliveGreen}x=3\sec u}$ , so that ${\color {OrangeRed}dx=3\sec u\tan u\,du}$ , gives ${\begin{aligned}\int {\frac {\color {OrangeRed}dx}{{\color {OliveGreen}x}^{2}{\sqrt {{\color {OliveGreen}x}^{2}-9}}}}&={\frac {1}{9}}\int {\frac {\color {OrangeRed}3\sec u\tan u\ du}{({\color {OliveGreen}3\sec u})^{2}{\sqrt {({\color {OliveGreen}3\sec u})^{2}-1}}}}\\&=\int {\frac {3\sec u\tan u\ du}{9\sec ^{2}u{\sqrt {9\sec ^{2}u-9}}}}\\{\rm {Recall\ that}}\ \sec ^{2}u-1=tan^{2}u,\\{\rm {thus\ simplifying\ the\ integrand\ gives}}\\&={\frac {1}{9}}\int {\frac {du}{\sec u}}\\&={\frac {1}{9}}\int \cos u\ du={\frac {1}{9}}\sin u+C.\end{aligned}}$ Now we must write the answer in terms of $x$ : First we know $\cos u={\frac {1}{\sec u}}$ And using the right triangle: ${\begin{aligned}\sin u={\sqrt {1-\cos ^{2}u}}&={\sqrt {1-{\frac {1}{\sec ^{2}u}}}}\\&={\sqrt {1-({\frac {3}{x}})^{2}}}\\&={\frac {\sqrt {x^{2}-9}}{x}}\end{aligned}}$ so: $\int {\frac {dx}{x^{2}{\sqrt {x^{2}-9}}}}={\color {blue}{\frac {\sqrt {x^{2}-9}}{9x}}+C}\left(={\frac {1}{9}}{\sqrt {1-\left({\frac {3}{x}}\right)^{\!2}}}+C\right).$