`int_(0)^(pi//2) sqrt(1- cos 2x) dx`

Answer

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

### Step 1: Use the trigonometric identity
We know that \( \cos 2x = 1 - 2\sin^2 x \). Therefore, we can rewrite \( 1 - \cos 2x \) as:
\[ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \]

### Step 2: Substitute into the integral
Now, substituting this back into the integral, we have:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \cos 2x} \, dx = \int_{0}^{\frac{\pi}{2}} \sqrt{2\sin^2 x} \, dx \]

### Step 3: Simplify the square root
The square root of \( 2\sin^2 x \) can be simplified:
\[ \sqrt{2\sin^2 x} = \sqrt{2} \cdot \sin x \]
Thus, the integral becomes:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{2} \cdot \sin x \, dx \]

### Step 4: Factor out the constant
We can factor out the constant \( \sqrt{2} \):
\[ \sqrt{2} \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 5: Integrate \( \sin x \)
The integral of \( \sin x \) is:
\[ \int \sin x \, dx = -\cos x \]
So, we evaluate:
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[-\cos x\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = -0 + 1 = 1 \]

### Step 6: Multiply by the constant
Now, substituting back, we have:
\[ \sqrt{2} \cdot 1 = \sqrt{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \cos 2x} \, dx = \sqrt{2} \]

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