$$ \int_0^{\frac{3\pi}{4}}\!\sqrt{1-\sin2x}\,\mathrm{d}x $$

To solve the integral $ \int_0^{\frac{3\pi}{4}} \sqrt{1 - \sin 2x} \, \mathrm{d}x $, we'll use trigonometric identities and substitution.

### Step 1: Simplify the integrand
First, recall the trigonometric identity:
\[
\sin 2x = 2 \sin x \cos x
\]
Thus, the integrand becomes:
\[
\sqrt{1 - \sin 2x}
\]

### Step 2: Use the identity for $1 - \sin 2x$
We can use the identity:
\[
1 - \sin 2x = \cos^2\left(x + \frac{\pi}{4}\right)
\]
Thus, the integral becomes:
\[
\int_0^{\frac{3\pi}{4}} \sqrt{\cos^2\left(x + \frac{\pi}{4}\right)} \, \mathrm{d}x
\]

### Step 3: Simplify the square root
Since $ \sqrt{\cos^2 y} = |\cos y| $, we get:
\[
\int_0^{\frac{3\pi}{4}} |\cos\left(x + \frac{\pi}{4}\right)| \, \mathrm{d}x
\]
Here, the absolute value is crucial because the cosine function can be negative over some intervals.

### Step 4: Evaluate the integral
To handle the absolute value, we must break the integral into intervals where $ \cos\left(x + \frac{\pi}{4}\right) $ is non-negative and where it is negative.

The function $ \cos\left(x + \frac{\pi}{4}\right) $ is non-negative in the interval $[0, \frac{3\pi}{4} - \frac{\pi}{4}]$, that is, $[0, \frac{\pi}{2}]$, and negative in $[\frac{\pi}{2}, \frac{3\pi}{4}]$.

So, the integral splits into two parts:
\[
\int_0^{\frac{\pi}{2}} \cos\left(x + \frac{\pi}{4}\right) \, \mathrm{d}x - \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \cos\left(x + \frac{\pi}{4}\right) \, \mathrm{d}x
\]

### Step 5: Integrate each part
1. For $ \int_0^{\frac{\pi}{2}} \cos\left(x + \frac{\pi}{4}\right) \, \mathrm{d}x $, use the substitution $ u = x + \frac{\pi}{4} $, $ \mathrm{d}u = \mathrm{d}x $:
   \[
   \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos u \, \mathrm{d}u = \sin u \Big|_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = \sin\frac{3\pi}{4} - \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0
   \]

2. For $ -\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \cos\left(x + \frac{\pi}{4}\right) \, \mathrm{d}x $:
   \[
   -\int_{\frac{3\pi}{4}}^{\pi} \cos u \, \mathrm{d}u = -\sin u \Big|_{\frac{3\pi}{4}}^{\pi} = -(\sin\pi - \sin\frac{3\pi}{4}) = 0 - \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}
   \]

### Final Answer
Thus, the integral evaluates to:
\[
\boxed{\frac{\sqrt{2}}{2}}
\]

Would you like further details on any step or have any questions?

---

1. What are the conditions where the cosine function is negative?
2. How does the identity $1 - \sin 2x = \cos^2(x + \frac{\pi}{4})$ derive?
3. Why is the integral split into two parts?
4. What is the significance of the absolute value in the integral?
5. How can substitution help simplify trigonometric integrals?

**Tip:** Always consider the sign of trigonometric functions when dealing with integrals that involve absolute values.

Learn how to evaluate the integral ∫₀^(3π/₄) √(1 - sin 2x) dx step-by-step using trigonometric identities and substitution. Understand why the integral splits into two parts and how to handle absolute values in integrals. Suitable for college-level students studying integral calculus and trigonometry.

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