MATH105 April 2015
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q6 (a) • Q6 (b) •
Question 02 (b)

Evaluate $\int _{2}^{4}{\frac {x^{2}4x+4}{\sqrt {12+4xx^{2}}}}\,dx$.

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

Try completing the square and making the substitution $u=x2$ (observe that the numerator then becomes $(x2)^{2}=u^{2}$).

Hint 2

Consider performing the trigonometric substitution $u=4\sin(t)$, and use the Pythagorean identity $\sin ^{2}(t)+\cos ^{2}(t)=1$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution

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Following the hints, we first complete the squares in the numerator and denominator:
 $\int _{2}^{4}{\frac {x^{2}4x+4}{\sqrt {12+4xx^{2}}}}\,dx=\int _{2}^{4}{\frac {(x2)^{2}}{\sqrt {(x2)^{2}+16}}}\,dx.$
We then make the substitution $u=x2\implies du=dx$:
 $\int _{2}^{4}{\frac {(x2)^{2}}{\sqrt {(x2)^{2}+16}}}\,dx=\int _{0}^{2}{\frac {u^{2}}{\sqrt {16u^{2}}}}\,du.$
Next, we make the trigonometric substitution $u=4\sin(t)\implies du=4\cos(t)\,dt,\ t=\arcsin(u/4)$:
 $\int _{0}^{2}{\frac {u^{2}}{\sqrt {16u^{2}}}}\,du=\int _{\arcsin(0)}^{\arcsin({\frac {1}{2}})}{\frac {16\sin ^{2}(t)\cdot 4\cos(t)}{\sqrt {1616\sin ^{2}(t)}}}\,dt.$
By the Pythagorean identity, $1616\sin ^{2}(t)=16(1\sin ^{2}(t))=16\cos ^{2}(t)$, so
 $\int _{\arcsin(0)}^{\arcsin({\frac {1}{2}})}{\frac {16\sin ^{2}(t)\cdot 4\cos(t)}{\sqrt {1616\sin ^{2}(t)}}}\,dt=\int _{0}^{\frac {\pi }{6}}{\frac {16\sin ^{2}(t)\cdot 4\cos(t)}{4\cos(t)}}\,dt=16\int _{0}^{\frac {\pi }{6}}\sin ^{2}(t)\,dt.$
Using the identity $\sin ^{2}(t)={\frac {1\cos(2t)}{2}}$, we obtain
 ${\begin{aligned}8\int _{0}^{\frac {\pi }{6}}1\cos(2t)\,dt&=8\int _{0}^{\frac {\pi }{6}}dt8\int _{0}^{\frac {\pi }{6}}\cos(2t)\,dt\\&=8\left(\left.t\right_{0}^{\frac {\pi }{6}}\right)8\left(\left.{\frac {\sin(2t)}{2}}\right_{0}^{\frac {\pi }{6}}\right)\\&={\frac {4\pi }{3}}4\sin \left({\frac {\pi }{3}}\right)\\&={\color {blue}{{\frac {4\pi }{3}}2{\sqrt {3}}}}.\end{aligned}}$
