`int(sin^(6)x+cos^(6)x+3sin^(2)x cos^(2)x)dx` is equal to

Options

A. `x+C`
B. `(3)/(2)sin2x+C`
C. `-(3)/(2)cos 2x+C`
D. `(1)/(3)sin 3x -cos 3x+C`

Explanation

To solve the integral \( \int (\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x) \, dx \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite \( \sin^6 x + \cos^6 x \) using the identity for the sum of cubes:
\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^6 x + \cos^6 x = 1 \cdot \left((\sin^2 x)^2 + (\cos^2 x)^2 - \sin^2 x \cos^2 x\right) \]

### Step 2: Simplify the expression
Now, we know that:
\[ (\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x \]
Thus, we can rewrite:
\[ \sin^6 x + \cos^6 x = (1 - 2 \sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3 \sin^2 x \cos^2 x \]

### Step 3: Substitute back into the integral
Now substituting back into the integral, we have:
\[ \int (1 - 3 \sin^2 x \cos^2 x + 3 \sin^2 x \cos^2 x) \, dx = \int (1) \, dx \]

### Step 4: Integrate
The integral simplifies to:
\[ \int 1 \, dx = x + C \]

### Final Answer
Thus, the solution to the integral is:
\[ \int (\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x) \, dx = x + C \]