Let `g(x)=int(1+2cosx)/((cosx+2)^2)dxa n dg(0)=0.` then the value of `8g(pi/2)` is __________

Explanation

To solve the problem, we need to evaluate the integral defined by the function \( g(x) \) and then find the value of \( 8g\left(\frac{\pi}{2}\right) \).

### Step-by-Step Solution:

1. **Define the function \( g(x) \)**:
\[ g(x) = \int \frac{1 + 2 \cos x}{(\cos x + 2)^2} \, dx \]

2. **Rewrite the integrand**:
We can separate the integrand:
\[ g(x) = \int \frac{1}{(\cos x + 2)^2} \, dx + \int \frac{2 \cos x}{(\cos x + 2)^2} \, dx \]

3. **Integrate the first term**:
Let \( u = \cos x + 2 \), then \( du = -\sin x \, dx \). The limits change accordingly when substituting, but we will focus on the indefinite integral for now:
\[ \int \frac{1}{u^2} (-du) = -\frac{1}{u} = -\frac{1}{\cos x + 2} \]

4. **Integrate the second term using integration by parts**:
For the second term \( \int \frac{2 \cos x}{(\cos x + 2)^2} \, dx \), we can use integration by parts:
- Let \( u = \frac{1}{\cos x + 2} \) and \( dv = 2 \cos x \, dx \).
- Then \( du = -\frac{\sin x}{(\cos x + 2)^2} \, dx \) and \( v = 2 \sin x \).

Applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives:
\[ \frac{2 \sin x}{\cos x + 2} - \int 2 \sin x \left(-\frac{\sin x}{(\cos x + 2)^2}\right) \, dx \]

5. **Combine results**:
After performing the integration by parts, we can combine the results:
\[ g(x) = -\frac{1}{\cos x + 2} + \frac{2 \sin x}{\cos x + 2} + C \]

6. **Determine the constant \( C \)**:
Given that \( g(0) = 0 \):
\[ g(0) = -\frac{1}{\cos(0) + 2} + \frac{2 \sin(0)}{\cos(0) + 2} + C = 0 \]
\[ -\frac{1}{1 + 2} + 0 + C = 0 \implies C = \frac{1}{3} \]

7. **Final expression for \( g(x) \)**:
\[ g(x) = -\frac{1}{\cos x + 2} + \frac{2 \sin x}{\cos x + 2} + \frac{1}{3} \]

8. **Evaluate \( g\left(\frac{\pi}{2}\right) \)**:
\[ g\left(\frac{\pi}{2}\right) = -\frac{1}{\cos\left(\frac{\pi}{2}\right) + 2} + \frac{2 \sin\left(\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{2}\right) + 2} + \frac{1}{3} \]
\[ = -\frac{1}{0 + 2} + \frac{2 \cdot 1}{0 + 2} + \frac{1}{3} = -\frac{1}{2} + 1 + \frac{1}{3} \]
\[ = \frac{2}{2} - \frac{1}{2} + \frac{1}{3} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

9. **Calculate \( 8g\left(\frac{\pi}{2}\right) \)**:
\[ 8g\left(\frac{\pi}{2}\right) = 8 \cdot \frac{5}{6} = \frac{40}{6} = \frac{20}{3} \]

Thus, the final answer is:
\[ \boxed{\frac{20}{3}} \]