3. Find the area between the curve $$ y=2^{\sin x} \cos x $$ and the $$ x $$-axis within the interval $$ [0,2 \pi] $$.

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To find the area between the curve $$ y = 2^{\sin x} \cos x $$ and the $$ x $$-axis over the interval $$ [0, 2\pi] $$, we need to evaluate the integral of the absolute value of the function:

$$
\text{Area} = \int_{0}^{2\pi} \left| 2^{\sin x} \cos x \right| dx
$$

### Step 1: Analyze the Function

The function $$ y = 2^{\sin x} \cos x $$ can take both positive and negative values depending on the value of $$ \cos x $$:
- $$ \cos x > 0 $$ when $$ x \in [0, \frac{\pi}{2}) $$ and $$ x \in [\frac{3\pi}{2}, 2\pi) $$
- $$ \cos x < 0 $$ when $$ x \in (\frac{\pi}{2}, \frac{3\pi}{2}) $$

### Step 2: Break the Integral into Intervals

We can split the integral into three parts where the function is positive or negative:

$$
\text{Area} = \int_{0}^{\frac{\pi}{2}} 2^{\sin x} \cos x \, dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left| 2^{\sin x} \cos x \right| dx + \int_{\frac{3\pi}{2}}^{2\pi} 2^{\sin x} \cos x \, dx
$$

### Step 3: Utilize Substitution

Let $$ u = \sin x $$. Then $$ du = \cos x \, dx $$.

For each interval:
1. **From $$ 0 $$ to $$ \frac{\pi}{2} $$:**
   $$
   \int_{0}^{\frac{\pi}{2}} 2^{\sin x} \cos x \, dx = \int_{0}^{1} 2^u \, du = \frac{2^u}{\ln 2}\Big|_{0}^{1} = \frac{2 - 1}{\ln 2} = \frac{1}{\ln 2}
   $$
   
2. **From $$ \frac{\pi}{2} $$ to $$ \frac{3\pi}{2} $$:**
   $$
   \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left| 2^{\sin x} \cos x \right| dx = \int_{1}^{-1} 2^u \, (-du) = \int_{-1}^{1} 2^u \, du = \frac{2^u}{\ln 2}\Big|_{-1}^{1} = \frac{2 - \frac{1}{2}}{\ln 2} = \frac{3}{2\ln 2}
   $$
   
3. **From $$ \frac{3\pi}{2} $$ to $$ 2\pi $$:**
   $$
   \int_{\frac{3\pi}{2}}^{2\pi} 2^{\sin x} \cos x \, dx = \int_{-1}^{0} 2^u \, du = \frac{2^u}{\ln 2}\Big|_{-1}^{0} = \frac{1 - \frac{1}{2}}{\ln 2} = \frac{1}{2\ln 2}
   $$

### Step 4: Sum the Results

Adding the three integrals together:

$$
\text{Area} = \frac{1}{\ln 2} + \frac{3}{2\ln 2} + \frac{1}{2\ln 2} = \frac{3}{\ln 2}
$$

### Final Answer

$$
\boxed{\dfrac{3}{\,\ln\,2\,}}
$$
The area between the curve $$ y = 2^{\sin x} \cos x $$ and the $$ x $$-axis from $$ 0 $$ to $$ 2\pi $$ is $$ \frac{3}{\ln 2} $$.

Find the area between the curve and the -axis within the
interval .

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Find the area between the curve and the -axis within the interval .