Science:Math Exam Resources/Courses/MATH105/April 2015/Question 02 (b)/Solution 1

Following the hints, we first complete the squares in the numerator and denominator:

$\int _{2}^{4}{\frac {x^{2}-4x+4}{\sqrt {12+4x-x^{2}}}}\,dx=\int _{2}^{4}{\frac {(x-2)^{2}}{\sqrt {-(x-2)^{2}+16}}}\,dx.$

We then make the substitution $u=x-2\implies du=dx$:

$\int _{2}^{4}{\frac {(x-2)^{2}}{\sqrt {-(x-2)^{2}+16}}}\,dx=\int _{0}^{2}{\frac {u^{2}}{\sqrt {16-u^{2}}}}\,du.$

Next, we make the trigonometric substitution $u=4\sin(t)\implies du=4\cos(t)\,dt,\ t=\arcsin(u/4)$:

$\int _{0}^{2}{\frac {u^{2}}{\sqrt {16-u^{2}}}}\,du=\int _{\arcsin(0)}^{\arcsin({\frac {1}{2}})}{\frac {16\sin ^{2}(t)\cdot 4\cos(t)}{\sqrt {16-16\sin ^{2}(t)}}}\,dt.$

By the Pythagorean identity, $16-16\sin ^{2}(t)=16(1-\sin ^{2}(t))=16\cos ^{2}(t)$, so

$\int _{\arcsin(0)}^{\arcsin({\frac {1}{2}})}{\frac {16\sin ^{2}(t)\cdot 4\cos(t)}{\sqrt {16-16\sin ^{2}(t)}}}\,dt=\int _{0}^{\frac {\pi }{6}}{\frac {16\sin ^{2}(t)\cdot 4\cos(t)}{4\cos(t)}}\,dt=16\int _{0}^{\frac {\pi }{6}}\sin ^{2}(t)\,dt.$

Using the identity $\sin ^{2}(t)={\frac {1-\cos(2t)}{2}}$, we obtain

${\begin{aligned}8\int _{0}^{\frac {\pi }{6}}1-\cos(2t)\,dt&=8\int _{0}^{\frac {\pi }{6}}dt-8\int _{0}^{\frac {\pi }{6}}\cos(2t)\,dt\\&=8\left(\left.t\right|_{0}^{\frac {\pi }{6}}\right)-8\left(\left.{\frac {\sin(2t)}{2}}\right|_{0}^{\frac {\pi }{6}}\right)\\&={\frac {4\pi }{3}}-4\sin \left({\frac {\pi }{3}}\right)\\&={\color {blue}{{\frac {4\pi }{3}}-2{\sqrt {3}}}}.\end{aligned}}$