`int(sin^(6)theta+cos^(6)theta)/(sin^(2)theta cos^(2)theta)d theta=`

Options

A. `tan theta+cot theta+3theta+c`
B. `tan theta-cot theta-3theta+c`
C. `tan theta+cot theta-3theta+c`
D. `tan theta-cot theta+3theta+c`

Answer

To solve the integral

\[ I = \int \frac{\sin^6 \theta + \cos^6 \theta}{\sin^2 \theta \cos^2 \theta} \, d\theta, \]

we can start by simplifying the expression in the integrand.

### Step 1: Rewrite \(\sin^6 \theta + \cos^6 \theta\)

Using the identity for the sum of cubes, we can express \(\sin^6 \theta + \cos^6 \theta\) as follows:

\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta). \]

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

\[ \sin^6 \theta + \cos^6 \theta = \sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta. \]

### Step 2: Express \(\sin^4 \theta + \cos^4 \theta\)

Next, we can use the identity:

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

Substituting this back, we get:

\[ \sin^6 \theta + \cos^6 \theta = (1 - 2\sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta = 1 - 3\sin^2 \theta \cos^2 \theta. \]

### Step 3: Substitute back into the integral

Now substituting this into our integral \(I\):

\[ I = \int \frac{1 - 3\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \, d\theta. \]

This can be separated into two integrals:

\[ I = \int \frac{1}{\sin^2 \theta \cos^2 \theta} \, d\theta - 3 \int d\theta. \]

### Step 4: Simplify the first integral

The first integral can be rewritten using the identity:

\[ \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} = \csc^2 \theta \sec^2 \theta. \]

Thus,

\[ \int \csc^2 \theta \sec^2 \theta \, d\theta. \]

### Step 5: Solve the integrals

The integral of \(\csc^2 \theta\) is \(-\cot \theta\) and the integral of \(\sec^2 \theta\) is \(\tan \theta\). Therefore, we can write:

\[ I = \int \csc^2 \theta \sec^2 \theta \, d\theta - 3\theta + C. \]

### Step 6: Final expression

Thus, the final expression for the integral is:

\[ I = -\cot \theta \tan \theta - 3\theta + C. \]