The area of the region bounded by `y=2cosx, y=3tanx` and the y-axis is

Options

A. `1+(3)/(2) log 3-log2`
B. `"log"(3)/(2)`
C. `1+3"log"(2)/(sqrt(3))`
D. `1+(3)/(2)log3-log8`

Answer

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 points of intersection
We need to find where the curves \( y = 2 \cos x \) and \( y = 3 \tan x \) intersect. This means we need to solve the equation:
\[ 2 \cos x = 3 \tan x \]
Rearranging gives:
\[ 2 \cos x = 3 \frac{\sin x}{\cos x} \]
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 \( \cos^2 x \) as \( 1 - \sin^2 x \):
\[ 2(1 - \sin^2 x) = 3 \sin x \]
Rearranging gives:
\[ 2 - 2 \sin^2 x - 3 \sin x = 0 \]
This is a quadratic equation in \( \sin x \).

### Step 2: Solve the quadratic equation
Let \( u = \sin x \). The equation becomes:
\[ 2u^2 + 3u - 2 = 0 \]
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ u = \frac{-3 \pm \sqrt{9 + 16}}{4} \]
\[ u = \frac{-3 \pm 5}{4} \]
Calculating the two possible values:
1. \( u = \frac{2}{4} = \frac{1}{2} \) (valid)
2. \( u = \frac{-8}{4} = -2 \) (not valid since \( u = \sin x \) must be between -1 and 1)

Thus, \( \sin x = \frac{1}{2} \) 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
First, we compute the integral of \( 2 \cos x \):
\[ \int 2 \cos x \, dx = 2 \sin x \]
Next, we compute the integral of \( 3 \tan x \):
\[ \int 3 \tan x \, dx = 3 \ln |\sec x| + C \]

Thus, the area becomes:
\[ A = \left[ 2 \sin x - 3 \ln |\sec x| \right]_0^{\frac{\pi}{6}} \]

### Step 5: Evaluate the integral at the bounds
Evaluate at \( x = \frac{\pi}{6} \):
\[ 2 \sin\left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1 \]
\[ 3 \ln |\sec\left(\frac{\pi}{6}\right)| = 3 \ln \left|\frac{2}{\sqrt{3}}\right| = 3 \left(\ln 2 - \frac{1}{2} \ln 3\right) \]

Now, evaluate at \( x = 0 \):
\[ 2 \sin(0) = 0 \]
\[ 3 \ln |\sec(0)| = 3 \ln 1 = 0 \]

Thus, the area is:
\[ A = \left(1 - 3 \left(\ln 2 - \frac{1}{2} \ln 3\right)\right) - 0 = 1 - 3 \ln 2 + \frac{3}{2} \ln 3 \]

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