Question

Compute the area of the curvilinear triangle bounded by the $y$-axis and the curves, $y=\tan x$ and $y=\frac{2}{3}\cos x$.

Answer 1

Step 1. Find the point of intersection of the curves $y=\tan x$ and $y=\frac{2}{3}\cos x$.

The enclosed area by the given curve is shown below,

Equate the given curves to find their point of intersection:

$$\begin{gathered} \therefore \tan x=\frac{2}{3}\cos x \\ \Rightarrow 3\tan x=2\cos x \\ \Rightarrow 3\sin x=2{\cos }^{2}x \\ \Rightarrow 3\sin x=2-2{\sin }^{2}x \\ \Rightarrow 2{\sin }^{2}x+3\sin x-2=0 \\ \Rightarrow \sin x=\frac{-3\pm \sqrt{9+16}}{4} \\ \Rightarrow \sin x=\frac{-3\pm 5}{4} \\ \Rightarrow \sin x=-2[\mathrm{not}\mathrm{possible}\because -1\le \sin x\le 1] \\ \Rightarrow \sin x=\frac{1}{2} \\ \Rightarrow x=\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6} \end{gathered}$$

Step 2: Calculate the area bounded by the $\mathrm{y}-\mathrm{axis}$ and the given curves.

Let $A$ be the area bounded by the $y$-axis and the given curves.

Therefore,

$\begin{array}{rcl}A& =& \left|{\int }_0^{\frac{\mathrm{\pi }}{6}}\left(\tan x-\frac{2}{3}\cos x\right)dx\right|\\ & =& {\left|\ln \left|\sec \mathrm{x}\right|-\frac{2}{3}\sin x\right|}_0^{\frac{\mathrm{\pi }}{6}}\\ & =& \left|\ln \left|\frac{2}{\sqrt{3}}\right|-\frac{2}{3}\times \frac{1}{2}-\ln \left|1\right|+\frac{2}{3}\times 0\right|\\ & =& \left|\ln \left|\frac{2}{\sqrt{3}}\right|-\frac{1}{3}\right|\\ & =& \frac{1}{3}-\ln \left|\frac{2}{\sqrt{3}}\right|\\ & =& \frac{1}{3}+\ln \left|\frac{\sqrt{3}}{2}\right|\mathrm{sq}.\mathrm{units}\end{array}$

Hence, the required area is $\frac{1}{3}+\ln \left|\frac{\sqrt{3}}{2}\right|\mathrm{sq}.\mathrm{units}$.