Let $P(x,\ y)\ (x>0,\ y>0)$ be a point on the curve $C: x^2-y^2=1$. If $x=\frac{e^u+e^{-u}}{2}\ (u\geq 0)$, then find the area bounded by the line $OP$, the $x$ axis and the curve $C$ in terms of $u$.
1. **Identify the curve and the point \( P(x, y) \):**
   The curve is given by \( C: x^2 - y^2 = 1 \). The point \( P(x, y) \) lies on this curve, and we are given \( x = \frac{e^u + e^{-u}}{2} \) for \( u \geq 0 \).

2. **Express \( y \) in terms of \( u \):**
   Since \( P(x, y) \) lies on the curve \( x^2 - y^2 = 1 \), we can substitute \( x = \frac{e^u + e^{-u}}{2} \) into the equation:
   \[
   \left( \frac{e^u + e^{-u}}{2} \right)^2 - y^2 = 1
   \]
   Simplifying the left-hand side:
   \[
   \frac{(e^u + e^{-u})^2}{4} - y^2 = 1
   \]
   \[
   \frac{e^{2u} + 2 + e^{-2u}}{4} - y^2 = 1
   \]
   \[
   \frac{e^{2u} + 2 + e^{-2u}}{4} - y^2 = 1
   \]
   \[
   \frac{e^{2u} + 2 + e^{-2u}}{4} - y^2 = 1
   \]
   \[
   y^2 = \frac{e^{2u} + 2 + e^{-2u}}{4} - 1
   \]
   \[
   y^2 = \frac{e^{2u} + 2 + e^{-2u} - 4}{4}
   \]
   \[
   y^2 = \frac{e^{2u} + e^{-2u} - 2}{4}
   \]
   \[
   y = \frac{e^u - e^{-u}}{2}
   \]

3. **Calculate the area bounded by the line \( OP \), the \( x \)-axis, and the curve \( C \):**
   The area can be divided into two parts: the area of the triangle \( \triangle OPH \) and the area under the curve from \( x = 1 \) to \( x = \frac{e^u + e^{-u}}{2} \).

   - **Area of \( \triangle OPH \):**
     The coordinates of \( P \) are \( \left( \frac{e^u + e^{-u}}{2}, \frac{e^u - e^{-u}}{2} \right) \). The area of the triangle is:
     \[
     \text{Area of } \triangle OPH = \frac{1}{2} \cdot x \cdot y = \frac{1}{2} \cdot \frac{e^u + e^{-u}}{2} \cdot \frac{e^u - e^{-u}}{2} = \frac{1}{8} (e^{2u} - e^{-2u})
     \]

   - **Area under the curve \( C \) from \( x = 1 \) to \( x = \frac{e^u + e^{-u}}{2} \):**
     The integral of \( y \) with respect to \( x \) from 1 to \( \frac{e^u + e^{-u}}{2} \):
     \[
     \int_1^{\frac{e^u + e^{-u}}{2}} y \, dx = \int_0^u \frac{e^u - e^{-u}}{2} \cdot \frac{d}{du} \left( \frac{e^u + e^{-u}}{2} \right) du
     \]
     \[
     = \int_0^u \frac{e^u - e^{-u}}{2} \cdot \frac{e^u - e^{-u}}{2} du
     \]
     \[
     = \frac{1}{4} \int_0^u (e^{2u} + e^{-2u} - 2) du
     \]
     \[
     = \frac{1}{4} \left[ \frac{e^{2u}}{2} - \frac{e^{-2u}}{2} - 2u \right]_0^u
     \]
     \[
     = \frac{1}{4} \left( \frac{e^{2u}}{2} - \frac{e^{-2u}}{2} - 2u - \left( \frac{1}{2} - \frac{1}{2} - 0 \right) \right)
     \]
     \[
     = \frac{1}{4} \left( \frac{e^{2u}}{2} - \frac{e^{-2u}}{2} - 2u \right)
     \]
     \[
     = \frac{1}{8} (e^{2u} - e^{-2u}) - \frac{1}{2} u
     \]

4. **Combine the areas:**
   The total area is the area of the triangle minus the area under the curve:
   \[
   \text{Total Area} = \frac{1}{8} (e^{2u} - e^{-2u}) - \left( \frac{1}{8} (e^{2u} - e^{-2u}) - \frac{1}{2} u \right)
   \]
   \[
   = \frac{1}{2} u
   \]

The final answer is \(\boxed{\frac{1}{2} u}\)