Let `f:[0,(pi)/(2)]toR` be continuous and satisfy `f'(x)=(1)/(1+cosx)` for all `x in(0,(pi)/(2))`. If f(0)=3 then `f((pi)/(2))` has the value equal to :

Answer

To solve the problem step by step, we start with the information given:

1. **Given Information**:
- The function \( f: [0, \frac{\pi}{2}] \to \mathbb{R} \) is continuous.
- The derivative \( f'(x) = \frac{1}{1 + \cos x} \) for all \( x \in (0, \frac{\pi}{2}) \).
- The initial condition \( f(0) = 3 \).

2. **Finding the Function \( f(x) \)**:
- To find \( f(x) \), we need to integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int \frac{1}{1 + \cos x} \, dx. \]

3. **Simplifying the Integral**:
- We can use the identity \( 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \):
\[ f(x) = \int \frac{1}{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) \, dx. \]

4. **Integrating**:
- The integral of \( \sec^2 u \) is \( \tan u \):
\[ f(x) = \frac{1}{2} \cdot 2 \tan \left(\frac{x}{2}\right) + C = \tan \left(\frac{x}{2}\right) + C. \]

5. **Using the Initial Condition**:
- We know \( f(0) = 3 \):
\[ f(0) = \tan(0) + C = 0 + C = 3. \]
- Therefore, \( C = 3 \).

6. **Final Expression for \( f(x) \)**:
- Thus, we have:
\[ f(x) = \tan \left(\frac{x}{2}\right) + 3. \]

7. **Finding \( f\left(\frac{\pi}{2}\right) \)**:
- Now, we need to find \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \tan \left(\frac{\frac{\pi}{2}}{2}\right) + 3 = \tan \left(\frac{\pi}{4}\right) + 3. \]
- Since \( \tan \left(\frac{\pi}{4}\right) = 1 \):
\[ f\left(\frac{\pi}{2}\right) = 1 + 3 = 4. \]

8. **Conclusion**:
- The value of \( f\left(\frac{\pi}{2}\right) \) is \( 4 \).