The area of the region whose boundaries are defined by the curves y=2 cos x, y=3 tan x, and the y-axis is

Options

A. `1+3" In "((2)/(sqrt(3)))` sq. units
B. `1+(3)/(2)" In "3-3" In "2` sq. units
C. `1+(3)/(2)" In "3-"In "2` sq. units
D. `" In "3-" In "2` sq. units

Explanation

To find the area of the region bounded by the curves \( y = 2 \cos x \), \( y = 3 \tan x \), and the y-axis, we will follow these steps:

### Step 1: Find the intersection points of the curves
To find the intersection points, we set the two equations equal to each other:
\[ 2 \cos x = 3 \tan x \]
We can rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \):
\[ 2 \cos x = 3 \frac{\sin x}{\cos x} \]
Multiplying both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ 2 \cos^2 x = 3 \sin x \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( \cos^2 x \) in terms of \( \sin x \):
\[ 2 (1 - \sin^2 x) = 3 \sin x \]
This simplifies to:
\[ 2 - 2 \sin^2 x = 3 \sin x \]
Rearranging gives us a quadratic equation:
\[ 2 \sin^2 x + 3 \sin x - 2 = 0 \]

### Step 2: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 3, c = -2 \):
\[ \sin x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ \sin x = \frac{-3 \pm \sqrt{9 + 16}}{4} \]
\[ \sin x = \frac{-3 \pm 5}{4} \]
Calculating the two possible values:
1. \( \sin x = \frac{2}{4} = \frac{1}{2} \) (valid)
2. \( \sin x = \frac{-8}{4} = -2 \) (not valid)

Thus, we have:
\[ \sin x = \frac{1}{2} \implies x = \frac{\pi}{6} \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{\pi}{6} \) is given by:
\[ A = \int_0^{\frac{\pi}{6}} (2 \cos x - 3 \tan x) \, dx \]

### Step 4: Compute the integral
We will compute the integral:
\[ A = \int_0^{\frac{\pi}{6}} 2 \cos x \, dx - \int_0^{\frac{\pi}{6}} 3 \tan x \, dx \]

Calculating the first integral:
\[ \int 2 \cos x \, dx = 2 \sin x \]
Evaluating from \( 0 \) to \( \frac{\pi}{6} \):
\[ 2 \sin\left(\frac{\pi}{6}\right) - 2 \sin(0) = 2 \cdot \frac{1}{2} - 0 = 1 \]

Calculating the second integral:
\[ \int 3 \tan x \, dx = 3 \ln |\sec x| + C \]
Evaluating from \( 0 \) to \( \frac{\pi}{6} \):
\[ 3 \ln |\sec(\frac{\pi}{6})| - 3 \ln |\sec(0)| \]
\[ = 3 \ln \left( \frac{2}{\sqrt{3}} \right) - 3 \ln(1) = 3 \ln \left( \frac{2}{\sqrt{3}} \right) = 3 \left( \ln 2 - \frac{1}{2} \ln 3 \right) \]

### Step 5: Combine results
Thus, the area is:
\[ A = 1 - 3 \left( \ln 2 - \frac{1}{2} \ln 3 \right) = 1 - 3 \ln 2 + \frac{3}{2} \ln 3 \]

### Final Answer
The area of the region is:
\[ A = 1 - 3 \ln 2 + \frac{3}{2} \ln 3 \text{ square units} \]