Science:Math Exam Resources/Courses/MATH101/April 2015/Question 01 (b) (ii)

MATH101 April 2015

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Question 01 (b) (ii)
For each integral, choose the substitution type that is most beneficial for evaluating the integral. (Write F, G, or H in each box; each answer will be used exactly once.) F. $x=a\sec \theta$ G. $x=a\sin \theta$ H: $x=a\tan \theta$ (ii) $\int {\frac {x^{4}-3}{\sqrt {1-4x^{2}}}}dx$

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 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
The denominator has the form ${\sqrt {a^{2}-x^{2}}}$, which trig identity help you simplify it?

Hint 2
Recall the trig identity: $1-\sin ^{2}\theta =\cos ^{2}\theta$ .

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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. Similar to part (i), first we need to get rid of the coefficient of $x$ by factoring. $\int {\frac {x^{4}-3}{\sqrt {1-4x^{2}}}}dx=\int {\frac {x^{4}-3}{\sqrt {4(1/4-x^{2})}}}dx=\int {\frac {x^{4}-3}{2{\sqrt {1/4-x^{2}}}}}dx=\int {\frac {x^{4}-3}{2{\sqrt {a^{2}-x^{2}}}}}dx$ Since the denominator has the form $a^{2}-x^{2}$, comparing that with the identity $1-\sin ^{2}\theta =\cos ^{2}\theta$, we conclude that the right choice of substitution is $x=a\sin \theta$, because then the denominator becomes ${\sqrt {a^{2}-a^{2}\sin ^{2}\theta }}={\sqrt {a^{2}(1-\sin ^{2}\theta )}}={\sqrt {a^{2}\cos ^{2}\theta }}=a|\cos \theta |$, which no longer involves a square root. The correct answer is then $\color {blue}G$.