If `U_(n)=int_(0)^(pi)(1-cosnx)/(1-cosx)dx` where `n` is positive integer of zero, then
The value of `U_(n)` is a. `pi//2` b. `pi` c. `npi//2` d. `npi`

Options

A. `pi//2`
B. `pi`
C. `npi//2`
D. `npi`

Explanation

To solve the problem, we need to evaluate the integral

\[ U_n = \int_0^{\pi} \frac{1 - \cos(nx)}{1 - \cos(x)} \, dx \]

where \( n \) is a positive integer.

### Step 1: Rewrite the Integral
We can rewrite the integral using the identity \( 1 - \cos(nx) = 2 \sin^2\left(\frac{nx}{2}\right) \) and \( 1 - \cos(x) = 2 \sin^2\left(\frac{x}{2}\right) \):

\[ U_n = \int_0^{\pi} \frac{2 \sin^2\left(\frac{nx}{2}\right)}{2 \sin^2\left(\frac{x}{2}\right)} \, dx = \int_0^{\pi} \frac{\sin^2\left(\frac{nx}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Use the Integral Property
Next, we can use the property of definite integrals. We can express \( U_n \) in terms of \( U_{n-1} \):

\[ U_n = U_{n-1} + \int_0^{\pi} \frac{\sin^2\left(\frac{nx}{2}\right) - \sin^2\left(\frac{(n-1)x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 3: Simplify the Difference of Sine Squares
Using the identity \( a^2 - b^2 = (a-b)(a+b) \):

\[ \sin^2\left(\frac{nx}{2}\right) - \sin^2\left(\frac{(n-1)x}{2}\right) = \left(\sin\left(\frac{nx}{2}\right) - \sin\left(\frac{(n-1)x}{2}\right)\right)\left(\sin\left(\frac{nx}{2}\right) + \sin\left(\frac{(n-1)x}{2}\right)\right) \]

### Step 4: Evaluate the Integral
The integral can be evaluated using the known result:

\[ \int_0^{\pi} \frac{\sin^2\left(\frac{nx}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} \, dx = n \cdot \pi \]

Thus, we find:

\[ U_n = n \cdot \frac{\pi}{2} \]

### Final Result
Therefore, the value of \( U_n \) is:

\[ U_n = \frac{n \pi}{2} \]

### Conclusion
The correct answer is option (c) \( \frac{n \pi}{2} \).

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