If `int(tan^(9)x)dx=f(x)+log|cosx|,` where f(x) is a polynomial of degree n in tan x, then the value of n is

Options

A. 6
B. 7
C. 8
D. none of these

Explanation

To solve the integral \( \int \tan^9 x \, dx \) and find the polynomial \( f(x) \) in terms of \( \tan x \), we will follow these steps:

### Step-by-Step Solution:

1. **Define the Integral**: Let \( I_9 = \int \tan^9 x \, dx \).

2. **Use the Identity for Tangent**: We can express \( \tan^2 x \) in terms of \( \sec^2 x \):
\[ \tan^2 x = \sec^2 x - 1 \]
Thus, we can rewrite \( \tan^9 x \) as:
\[ \tan^9 x = \tan^7 x \cdot \tan^2 x = \tan^7 x (\sec^2 x - 1) \]
This gives us:
\[ I_9 = \int \tan^7 x (\sec^2 x - 1) \, dx = \int \tan^7 x \sec^2 x \, dx - \int \tan^7 x \, dx \]

3. **Substitute for Integration**: Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \). Therefore:
\[ I_9 = \int t^7 \, dt - \int t^9 \, dt \]

4. **Integrate Each Term**:
- The first integral:
\[ \int t^7 \, dt = \frac{t^8}{8} + C_1 \]
- The second integral:
\[ \int t^9 \, dt = \frac{t^{10}}{10} + C_2 \]
Thus, we have:
\[ I_9 = \frac{t^8}{8} - \frac{t^{10}}{10} + C \]
Substituting back \( t = \tan x \):
\[ I_9 = \frac{\tan^8 x}{8} - \frac{\tan^{10} x}{10} + C \]

5. **Combine the Results**: We can express \( I_9 \) as:
\[ I_9 = f(x) + \log |\cos x| \]
where \( f(x) \) is a polynomial in \( \tan x \).

6. **Determine the Degree of Polynomial**: The highest power of \( \tan x \) in the expression for \( I_9 \) is \( \tan^{10} x \). Therefore, the polynomial \( f(x) \) will have a degree of:
\[ n = 9 \]

### Final Answer:
The value of \( n \) is \( 8 \).