The area bounded by the curve `y=cosx and y=sin 2x, AA x in [(pi)/(6), (pi)/(2)]` is equal to

Options

A. `(pi)/(2)" sq. units"`
B. `(pi)/(3)" sq. units"`
C. `7/4-(3)^(1/2)"sq. units"
D. `(1)/(4)" sq. units"`

Answer

To find the area bounded by the curves \( y = \cos x \) and \( y = \sin 2x \) over the interval \( x \in \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \), we can follow these steps:

### Step 1: Identify the curves and their intersection points
We need to find the points where the curves \( y = \cos x \) and \( y = \sin 2x \) intersect. We set them equal to each other:

\[ \cos x = \sin 2x \]

Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the equation as:

\[ \cos x = 2 \sin x \cos x \]

Rearranging gives:

\[ \cos x (1 - 2 \sin x) = 0 \]

This leads to two cases:

1. \( \cos x = 0 \)
2. \( 1 - 2 \sin x = 0 \) which gives \( \sin x = \frac{1}{2} \)

From the first case, \( \cos x = 0 \) occurs at \( x = \frac{\pi}{2} \).

From the second case, \( \sin x = \frac{1}{2} \) occurs at \( x = \frac{\pi}{6} \).

Thus, the intersection points in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \) are \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{2} \).

### Step 2: Determine which function is on top
We need to determine which function is greater in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \).

- At \( x = \frac{\pi}{6} \):
\[ y = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
\[ y = \sin\left(2 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

- At \( x = \frac{\pi}{3} \):
\[ y = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
\[ y = \sin\left(2 \cdot \frac{\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

- At \( x = \frac{\pi}{2} \):
\[ y = \cos\left(\frac{\pi}{2}\right) = 0 \]
\[ y = \sin\left(2 \cdot \frac{\pi}{2}\right) = \sin(\pi) = 0 \]

From this, we can see that \( \sin 2x \) is above \( \cos x \) in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \).

### Step 3: Set up the integral for the area
The area \( A \) bounded by the curves is given by:

\[ A = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin 2x - \cos x) \, dx \]

### Step 4: Compute the integral
Now we compute the integral:

\[ A = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin 2x \, dx - \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos x \, dx \]

Calculating each integral separately:

1. **Integral of \( \sin 2x \)**:
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \]
Evaluating from \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \):
\[ -\frac{1}{2} \left[ \cos(2 \cdot \frac{\pi}{2}) - \cos(2 \cdot \frac{\pi}{6}) \right] = -\frac{1}{2} \left[ \cos(\pi) - \cos\left(\frac{\pi}{3}\right) \right] \]
\[ = -\frac{1}{2} \left[ -1 - \frac{1}{2} \right] = -\frac{1}{2} \left[ -\frac{3}{2} \right] = \frac{3}{4} \]

2. **Integral of \( \cos x \)**:
\[ \int \cos x \, dx = \sin x \]
Evaluating from \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \):
\[ \left[ \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{6}\right) \right] = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 5: Combine the results
Now we combine the results of the integrals:

\[ A = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4} \]

### Final Answer
Thus, the area bounded by the curves \( y = \cos x \) and \( y = \sin 2x \) in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \) is:

\[ \boxed{\frac{1}{4}} \]