Question

Compute the area of the curvilinear triangle bounded by the y-axis & the curve, $y=\tan x$ & $y=\left(2/3\right)\cos x$

• A. $\frac{1}{3}+ln\left[\frac{\sqrt{3}}{2}\right]sq.units$
• B. $\frac{1}{3}-ln\left[\frac{\sqrt{3}}{2}\right]sq.units$
• C. $\frac{2}{3}+ln\left[\frac{\sqrt{3}}{2}\right]sq.units$
• D. $\frac{1}{3}+ln\left[\frac{\sqrt{1}}{2}\right]sq.units$

Answer 1

The correct option is A $\frac{1}{3}+ln\left[\frac{\sqrt{3}}{2}\right]sq.units$

For the point of intersection
$tan\left(x\right)=\frac{2}{3}cos\left(x\right)$
Or
$3tan\left(x\right)=2cos\left(x\right)$
$3sin\left(x\right)=2co{s}^2\left(x\right)$
Or
$3sin\left(x\right)=2-2si{n}^2\left(x\right)$
Or
$2si{n}^2\left(x\right)+3sin\left(x\right)-2=0$
Hence
$sin\left(x\right)=\frac{-3\pm \sqrt{9+16}}{4}$
$=\frac{-3\pm 5}{4}$
Hence
$sin\left(x\right)=-2$ ...(not possible) and $sin\left(x\right)=\frac{1}{2}$
Hence
$x=\frac{\pi }{6}$ , $x=\frac{5\pi }{6}$.
Hence the required area will be
$=|{\int }_0^{\frac{\pi }{6}}tan\left(x\right)-\frac{2}{3}cos\left(x\right)|$
$=ln|sec\left(x\right)|-\frac{2}{3}sin\left(x\right){|}_0^{\frac{\pi }{6}}$
$=|ln\left(\frac{2}{\sqrt{3}}\right)-\frac{1}{3}|$
$=\frac{1}{3}-ln\left(\frac{2}{\sqrt{3}}\right)$
$=\frac{1}{3}+ln|\frac{\sqrt{3}}{2}|$ sq units.