Evaluate `int(sin^(6)x)/(cos^(8)x)dx`

Explanation

To evaluate the integral \(\int \frac{\sin^6 x}{\cos^8 x} \, dx\), we can follow these steps:

### Step 1: Rewrite the integral
We can rewrite the integral in a more manageable form:
\[ \int \frac{\sin^6 x}{\cos^8 x} \, dx = \int \frac{\sin^6 x}{\cos^6 x} \cdot \frac{1}{\cos^2 x} \, dx \]
This can be expressed as:
\[ \int \tan^6 x \cdot \sec^2 x \, dx \]

### Step 2: Use substitution
Let \( t = \tan x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = \sec^2 x \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x} \]
Substituting \( t \) into the integral gives:
\[ \int t^6 \sec^2 x \cdot \frac{dt}{\sec^2 x} = \int t^6 \, dt \]

### Step 3: Integrate
Now we can integrate \( t^6 \):
\[ \int t^6 \, dt = \frac{t^{7}}{7} + C \]

### Step 4: Substitute back
Now, we substitute back \( t = \tan x \):
\[ \frac{\tan^7 x}{7} + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{\sin^6 x}{\cos^8 x} \, dx = \frac{\tan^7 x}{7} + C \]