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

Options

A. `(1)/(7)tan^(7)x+C`
B. `(1)/(7)sec^(7)x+C`
C. `log|cos^(6)x|+C`
D. none of these

Explanation

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

### Step 1: Rewrite the Integral
We start by rewriting 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
Next, we use the substitution:
\[ t = \tan x \quad \Rightarrow \quad dt = \sec^2 x \, dx \]
This means that \(dx = \frac{dt}{\sec^2 x}\). However, since we already have \(\sec^2 x \, dx\) in the integral, we can substitute directly:
\[ \int t^6 \, dt \]

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

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

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