`int_0^2sqrt((2+x)/(2-x))dx=`

Options

A. `pi+2`
B. `pi+3/2`
C. `pi+1`
D. `pi-1`

Answer

To solve the integral \( I = \int_0^2 \sqrt{\frac{2+x}{2-x}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^2 \sqrt{\frac{2+x}{2-x}} \, dx \]

### Step 2: Multiply and Divide by \(\sqrt{2+x}\)
We can simplify the integrand by multiplying and dividing by \(\sqrt{2+x}\):
\[ I = \int_0^2 \frac{\sqrt{(2+x)(2+x)}}{\sqrt{(2-x)(2+x)}} \, dx = \int_0^2 \frac{2+x}{\sqrt{(2-x)(2+x)}} \, dx \]

### Step 3: Substitute \( x = 2 \sin \theta \)
To simplify the integral further, we can use the substitution \( x = 2 \sin \theta \). Then, the differential \( dx \) becomes:
\[ dx = 2 \cos \theta \, d\theta \]
We also need to change the limits of integration:
- When \( x = 0 \), \( \sin \theta = 0 \) implies \( \theta = 0 \).
- When \( x = 2 \), \( \sin \theta = 1 \) implies \( \theta = \frac{\pi}{2} \).

### Step 4: Substitute into the Integral
Substituting \( x = 2 \sin \theta \) into the integral gives:
\[ I = \int_0^{\frac{\pi}{2}} \frac{2 + 2 \sin \theta}{\sqrt{(2 - 2\sin\theta)(2 + 2\sin\theta)}} \cdot 2 \cos \theta \, d\theta \]
This simplifies to:
\[ I = 2 \int_0^{\frac{\pi}{2}} \frac{1 + \sin \theta}{\sqrt{2(1 - \sin \theta)(1 + \sin \theta)}} \cdot 2 \cos \theta \, d\theta \]

### Step 5: Simplify the Denominator
The denominator can be simplified:
\[ \sqrt{2(1 - \sin \theta)(1 + \sin \theta)} = \sqrt{2(1 - \sin^2 \theta)} = \sqrt{2 \cos^2 \theta} = \sqrt{2} \cos \theta \]
Thus, we have:
\[ I = 2 \int_0^{\frac{\pi}{2}} \frac{1 + \sin \theta}{\sqrt{2} \cos \theta} \cdot 2 \cos \theta \, d\theta = \frac{4}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (1 + \sin \theta) \, d\theta \]

### Step 6: Evaluate the Integral
Now we can evaluate the integral:
\[ \int_0^{\frac{\pi}{2}} (1 + \sin \theta) \, d\theta = \int_0^{\frac{\pi}{2}} 1 \, d\theta + \int_0^{\frac{\pi}{2}} \sin \theta \, d\theta \]
Calculating these:
- \(\int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}\)
- \(\int_0^{\frac{\pi}{2}} \sin \theta \, d\theta = -\cos \theta \bigg|_0^{\frac{\pi}{2}} = 0 - (-1) = 1\)

Thus:
\[ \int_0^{\frac{\pi}{2}} (1 + \sin \theta) \, d\theta = \frac{\pi}{2} + 1 \]

### Step 7: Final Calculation
Substituting back, we find:
\[ I = \frac{4}{\sqrt{2}} \left( \frac{\pi}{2} + 1 \right) = 2\sqrt{2} \left( \frac{\pi}{2} + 1 \right) \]

### Final Result
Thus, the final result of the integral is:
\[ I = \pi + 2 \]