`" if " int (sin 2x- cos 2x) dx=(1)/(sqrt(2)) sin (2x-k)+c " then " k=?`

Answer

To solve the problem, we need to find the value of \( k \) in the equation:

\[ \int (\sin 2x - \cos 2x) \, dx = \frac{1}{\sqrt{2}} \sin(2x - k) + C \]

### Step 1: Calculate the integral

We start by calculating the integral on the left side:

\[ \int (\sin 2x - \cos 2x) \, dx = \int \sin 2x \, dx - \int \cos 2x \, dx \]

### Step 2: Integrate each term

Using the integration formulas:
- \(\int \sin(nx) \, dx = -\frac{1}{n} \cos(nx) + C\)
- \(\int \cos(nx) \, dx = \frac{1}{n} \sin(nx) + C\)

We can compute:

1. For \(\int \sin 2x \, dx\):
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C_1 \]

2. For \(\int \cos 2x \, dx\):
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x + C_2 \]

Thus, combining these results, we have:

\[ \int (\sin 2x - \cos 2x) \, dx = -\frac{1}{2} \cos 2x - \frac{1}{2} \sin 2x + C \]

### Step 3: Factor out \(-\frac{1}{2}\)

We can factor out \(-\frac{1}{2}\):

\[ = -\frac{1}{2} (\cos 2x + \sin 2x) + C \]

### Step 4: Rewrite in terms of sine

Next, we can express \(\cos 2x + \sin 2x\) in a single sine function using the identity:

\[ \cos A + \sin A = \sqrt{2} \sin\left(A + \frac{\pi}{4}\right) \]

Thus, we rewrite:

\[ \cos 2x + \sin 2x = \sqrt{2} \sin\left(2x + \frac{\pi}{4}\right) \]

Substituting this back gives:

\[ -\frac{1}{2} (\cos 2x + \sin 2x) = -\frac{1}{2} \sqrt{2} \sin\left(2x + \frac{\pi}{4}\right) \]

### Step 5: Simplify the expression

This can be simplified to:

\[ -\frac{1}{\sqrt{2}} \sin\left(2x + \frac{\pi}{4}\right) + C \]

### Step 6: Set equal to the right side

Now we equate this to the right side of the original equation:

\[ -\frac{1}{\sqrt{2}} \sin\left(2x + \frac{\pi}{4}\right) + C = \frac{1}{\sqrt{2}} \sin(2x - k) + C \]

### Step 7: Compare the two sides

Ignoring the constant \( C \), we have:

\[ -\sin\left(2x + \frac{\pi}{4}\right) = \sin(2x - k) \]

### Step 8: Use the sine identity

Using the identity \(\sin(A) = -\sin(B)\) implies:

\[ A = \pi - B \quad \text{or} \quad A = -B \]

Thus, we can write:

\[ 2x + \frac{\pi}{4} = \pi - (2x - k) \]

Solving this gives:

\[ 2x + \frac{\pi}{4} = \pi - 2x + k \]

### Step 9: Rearranging the equation

Rearranging, we find:

\[ 4x + \frac{\pi}{4} = \pi + k \]

### Step 10: Isolate \( k \)

To isolate \( k \), we can set \( x = 0 \):

\[ 0 + \frac{\pi}{4} = \pi + k \implies k = \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \]

Thus, we find:

\[ k = -\frac{3\pi}{4} \]

### Final Answer

The value of \( k \) is:

\[ \boxed{-\frac{3\pi}{4}} \]