MATH101 April 2008
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Question 03 (d)

FullSolution Problem. Justify your answers and show all your work. Simplification of answers is not required.
Evaluate the following integral.
 $\int {\frac {dx}{\sqrt {x^{2}+16}}}.$

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 a trigonometric substitution.

Hint 2

Your trigonometric substitution should make use of the fact that $a^{2}+a^{2}\tan ^{2}\theta =a^{2}\sec ^{2}\theta .$

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Solution

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We use the trigonometric substitution
 $x=4\tan \theta ,\quad dx=4\sec ^{2}\theta \,d\theta$
and obtain
${\begin{aligned}\int {\frac {dx}{\sqrt {x^{2}+16}}}&=\int {\frac {4\sec ^{2}\theta \,d\theta }{\sqrt {16\tan ^{2}\theta +16}}}\\&=\int {\frac {4\sec ^{2}\theta \,d\theta }{4\sec \theta }}\\&=\int \sec \theta \,d\theta .\end{aligned}}$
To solve this integral we substitute
 $u=\sec \theta +\tan \theta ,\quad du=(\tan \theta +\sec \theta )\sec \theta \,d\theta ,$
and obtain
${\begin{aligned}\int \sec \theta \,d\theta &=\int {\frac {\sec \theta \,du}{(\tan \theta +\sec \theta )\sec \theta }}\\&=\int {\frac {du}{u}}=\ln u\\&=\ln \sec \theta +\tan \theta +C.\end{aligned}}$
We use the first substitution $\tan \theta =x/4$ to find that $\cos \theta ={\frac {4}{\sqrt {x^{2}+16}}}$ so that our final answer becomes
 $\int {\frac {dx}{\sqrt {x^{2}+16}}}=\ln \left{\frac {\sqrt {x^{2}+16}}{4}}+{\frac {x}{4}}\right+C.$
Note that $\ln \left{\frac {\sqrt {x^{2}+16}}{4}}+{\frac {x}{4}}\right+C$ can also be written as $\sinh ^{1}\left({\frac {x}{4}}\right)+C,$ the inverse hyperbolic sine function.

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