Let `A_(n) = int tan^(n) xdx, AA n in N`. If `A_(10) + A_(12) = (tan^(m)x)/(m) + lambda` (where `lambda` is an arbitrary constant), then the value of m is equal to

Options

A. 10
B. 11
C. 12
D. 13

Answer

To solve the problem, we need to compute \( A_{10} + A_{12} \) where \( A_n = \int \tan^n x \, dx \). We will use integration techniques to find the value of \( m \) in the expression \( A_{10} + A_{12} = \frac{\tan^m x}{m} + \lambda \).

### Step-by-Step Solution:

1. **Define the Integrals**:
\[ A_{10} = \int \tan^{10} x \, dx, \quad A_{12} = \int \tan^{12} x \, dx \]

2. **Combine the Integrals**:
\[ A_{10} + A_{12} = \int \tan^{10} x \, dx + \int \tan^{12} x \, dx = \int (\tan^{10} x + \tan^{12} x) \, dx \]

3. **Factor Out \(\tan^{10} x\)**:
\[ A_{10} + A_{12} = \int \tan^{10} x (1 + \tan^2 x) \, dx \]

4. **Use the Identity**:
Since \( 1 + \tan^2 x = \sec^2 x \), we can rewrite the integral:
\[ A_{10} + A_{12} = \int \tan^{10} x \sec^2 x \, dx \]

5. **Substitution**:
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \), so:
\[ A_{10} + A_{12} = \int t^{10} \, dt \]

6. **Integrate**:
\[ \int t^{10} \, dt = \frac{t^{11}}{11} + \lambda = \frac{\tan^{11} x}{11} + \lambda \]

7. **Compare with Given Expression**:
We have:
\[ A_{10} + A_{12} = \frac{\tan^{11} x}{11} + \lambda \]
The problem states that:
\[ A_{10} + A_{12} = \frac{\tan^m x}{m} + \lambda \]

8. **Identify \( m \)**:
By comparing both expressions, we see that \( m = 11 \).

### Final Answer:
The value of \( m \) is \( 11 \).