Science:Math Exam Resources/Courses/MATH103/April 2016/Question 02 (c)/Solution 1

Let $u=\sin(3x),\,du=3\cos(3x)\,dx$.

Then the lower and upper limits of the integral become

$x={\frac {\pi }{12}}\implies u=\sin \left(3\cdot {\frac {\pi }{12}}\right)=\sin \left({\frac {\pi }{4}}\right)={\frac {\sqrt {2}}{2}},$

$x={\frac {\pi }{6}}\implies u=\sin \left(3\cdot {\frac {\pi }{6}}\right)=\sin \left({\frac {\pi }{2}}\right)=1.$

Hence

${\begin{aligned}\int _{\frac {\pi }{12}}^{\frac {\pi }{6}}{\frac {\cos(3x)}{\sin(3x)}}\,dx&=\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {du}{3u}}\\&=\left.{\frac {1}{3}}\ln x\right|_{\frac {\sqrt {2}}{2}}^{1}\\&=-{\frac {1}{3}}\ln \left({\frac {\sqrt {2}}{2}}\right)\\&=\color {blue}{\frac {\ln 2}{6}}.\end{aligned}}$