`int_(0)^((pi)/2)(x-[cosx])dx=`
(where `[t]=` greatest integer less or equal to `t`)

Options

A. `(pi)/4`
B. `(pi^(2))/8-(pi)/8`
C. `(pi^(2))/8-1`
D. `(pi^(2))/8`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \left( x - [\cos x] \right) dx \), where \([t]\) denotes the greatest integer less than or equal to \( t \), we will break down the problem step by step.

### Step 1: Understand the function \([\cos x]\)

The function \(\cos x\) varies between 1 and 0 for \(x\) in the interval \([0, \frac{\pi}{2}]\). Therefore, we can analyze the values of \([\cos x]\):
- For \(x = 0\), \(\cos(0) = 1\) so \([\cos(0)] = 1\).
- For \(x = \frac{\pi}{2}\), \(\cos\left(\frac{\pi}{2}\right) = 0\) so \([\cos\left(\frac{\pi}{2}\right)] = 0\).

Since \(\cos x\) decreases from 1 to 0 in the interval \([0, \frac{\pi}{2}]\), we can conclude:
- For \(0 \leq x < \frac{\pi}{3}\), \(\cos x\) is greater than or equal to \(\frac{1}{2}\) and thus \([\cos x] = 0\).
- For \(\frac{\pi}{3} \leq x < \frac{\pi}{2}\), \(\cos x\) is less than \(\frac{1}{2}\) and thus \([\cos x] = 0\).

### Step 2: Rewrite the integral

Given that \([\cos x] = 0\) for all \(x\) in the interval \([0, \frac{\pi}{2}]\), we can simplify the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( x - 0 \right) dx = \int_{0}^{\frac{\pi}{2}} x \, dx \]

### Step 3: Calculate the integral

Now we compute the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} x \, dx \]
Using the formula for the integral of \(x\):
\[ \int x \, dx = \frac{x^2}{2} + C \]
We evaluate it from 0 to \(\frac{\pi}{2}\):
\[ I = \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{\left(\frac{\pi}{2}\right)^2}{2} - \frac{0^2}{2} = \frac{\frac{\pi^2}{4}}{2} = \frac{\pi^2}{8} \]

### Final Answer

Thus, the value of the integral is:
\[ I = \frac{\pi^2}{8} \]

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