The value of `int_(0)^(1)cos^(-1)((x-x^(2))-sqrt((1-x^(2))(2x-x^(2))))dx` is equal to ___________.

Explanation

To solve the integral \( I = \int_{0}^{1} \cos^{-1} \left( x - x^2 - \sqrt{(1 - x^2)(2x - x^2)} \right) dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Let us denote the integral as:
\[ I = \int_{0}^{1} \cos^{-1} \left( x - x^2 - \sqrt{(1 - x^2)(2x - x^2)} \right) dx \]

### Step 2: Simplifying the Argument of the Cosine Inverse
We can rewrite the expression inside the cosine inverse:
\[ x - x^2 - \sqrt{(1 - x^2)(2x - x^2)} \]
We can factor out \( x(1-x) \) and simplify the square root term.

### Step 3: Use the Identity for Cosine Inverse
Using the identity:
\[ \cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}(AB - \sqrt{(1-A^2)(1-B^2)}) \]
where \( A = x \) and \( B = 1 - x \).

### Step 4: Split the Integral
Thus, we can express \( I \) as:
\[ I = \int_{0}^{1} \cos^{-1}(x) dx + \int_{0}^{1} \cos^{-1}(1-x) dx \]
Using the property of definite integrals, we know:
\[ \int_{0}^{1} \cos^{-1}(1-x) dx = \int_{0}^{1} \cos^{-1}(x) dx \]
So we can write:
\[ I = 2 \int_{0}^{1} \cos^{-1}(x) dx \]

### Step 5: Change of Variables
Now, we can change the variable by letting \( x = \cos(\theta) \):
\[ dx = -\sin(\theta) d\theta \]
The limits change from \( x = 0 \) to \( x = 1 \) which corresponds to \( \theta = \frac{\pi}{2} \) to \( \theta = 0 \).

### Step 6: Evaluate the Integral
Thus,
\[ I = 2 \int_{\frac{\pi}{2}}^{0} \theta (-\sin(\theta)) d\theta = 2 \int_{0}^{\frac{\pi}{2}} \theta \sin(\theta) d\theta \]

### Step 7: Integration by Parts
Using integration by parts where \( u = \theta \) and \( dv = \sin(\theta) d\theta \):
\[ du = d\theta, \quad v = -\cos(\theta) \]
Thus,
\[ \int \theta \sin(\theta) d\theta = -\theta \cos(\theta) + \int \cos(\theta) d\theta = -\theta \cos(\theta) + \sin(\theta) \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):
\[ = \left[-\theta \cos(\theta) + \sin(\theta)\right]_{0}^{\frac{\pi}{2}} = \left[-\frac{\pi}{2} \cdot 0 + 1\right] - \left[0 + 0\right] = 1 \]

### Step 8: Final Calculation
Thus,
\[ I = 2 \cdot 1 = 2 \]

The value of the integral is:
\[ \boxed{2} \]