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?

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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.