`int(sin^8x-cos^8x)/(1-2sin^2xcos^2x)dx=`

Options

A. `sin 2x + C`
B. ` (sin 2x)/(2) +C`
C. `(-sin 2x)/(2) +C`
D. `-2 sin 2x +C`

Answer

To solve the integral

\[ \int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the numerator using the difference of squares
We can express \(\sin^8 x - \cos^8 x\) as a difference of squares:

\[ \sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x). \]

### Step 2: Further simplify \(\sin^4 x - \cos^4 x\)
We can apply the difference of squares again to \(\sin^4 x - \cos^4 x\):

\[ \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x). \]

Since \(\sin^2 x + \cos^2 x = 1\), we have:

\[ \sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x. \]

### Step 3: Substitute back into the integral
Now we substitute this back into our integral:

\[ \int \frac{(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)}{1 - 2\sin^2 x \cos^2 x} \, dx. \]

### Step 4: Simplify the denominator
The denominator \(1 - 2\sin^2 x \cos^2 x\) can be rewritten using the identity:

\[ 1 - 2\sin^2 x \cos^2 x = \sin^2 x + \cos^2 x - 2\sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x. \]

### Step 5: Substitute the denominator
Now, substituting this back into the integral gives:

\[ \int \frac{(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 6: Cancel out terms
The \(\sin^4 x + \cos^4 x\) terms cancel out:

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

### Step 7: Rewrite the integrand
We can rewrite \(\sin^2 x - \cos^2 x\) using the double angle identity:

\[ \sin^2 x - \cos^2 x = -\cos 2x. \]

### Step 8: Integrate
Now we can integrate:

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

### Final Answer
Thus, the final answer is:

\[ -\frac{\sin 2x}{2} + C. \]

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