MATH101 April 2016
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q10 (a) • Q10 (b) • Q11 (a) • Q11 (b) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) •
Question 3 (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 $x=3\sec u$.

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.
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Solution

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Substituting ${\color {OliveGreen}x=3\sec u}$, so that ${\color {OrangeRed}dx=3\sec u\tan u\,du}$, gives
${\begin{aligned}\int {\frac {\color {OrangeRed}dx}{{\color {OliveGreen}x}^{2}{\sqrt {{\color {OliveGreen}x}^{2}9}}}}&={\frac {1}{9}}\int {\frac {\color {OrangeRed}3\sec u\tan u\ du}{({\color {OliveGreen}3\sec u})^{2}{\sqrt {({\color {OliveGreen}3\sec u})^{2}1}}}}\\&=\int {\frac {3\sec u\tan u\ du}{9\sec ^{2}u{\sqrt {9\sec ^{2}u9}}}}\\{\rm {Recall\ that}}\ \sec ^{2}u1=tan^{2}u,\\{\rm {thus\ simplifying\ the\ integrand\ gives}}\\&={\frac {1}{9}}\int {\frac {du}{\sec u}}\\&={\frac {1}{9}}\int \cos u\ du={\frac {1}{9}}\sin u+C.\end{aligned}}$
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:
$\int {\frac {dx}{x^{2}{\sqrt {x^{2}9}}}}={\color {blue}{\frac {\sqrt {x^{2}9}}{9x}}+C}\left(={\frac {1}{9}}{\sqrt {1\left({\frac {3}{x}}\right)^{\!2}}}+C\right).$
