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 evaluate the integrals \( A_{10} \) and \( A_{12} \) defined as:

\[ A_n = \int \tan^n x \, dx \]

We are given that:

\[ A_{10} + A_{12} = \frac{\tan^m x}{m} + \lambda \]

where \( \lambda \) is an arbitrary constant. We need to find the value of \( m \).

### Step 1: Evaluate \( A_{10} \) and \( A_{12} \)

Using integration by parts, we can derive a recursive relation for \( A_n \):

\[ A_n = \int \tan^n x \, dx = -\frac{1}{n} \tan^{n-1} x + \frac{n-1}{n} A_{n-2} \]

This gives us:

\[ A_{10} = -\frac{1}{10} \tan^9 x + \frac{9}{10} A_8 \]

\[ A_{12} = -\frac{1}{12} \tan^{11} x + \frac{11}{12} A_{10} \]

### Step 2: Substitute \( A_{10} \) into \( A_{12} \)

Now we substitute \( A_{10} \) into \( A_{12} \):

\[ A_{12} = -\frac{1}{12} \tan^{11} x + \frac{11}{12} \left( -\frac{1}{10} \tan^9 x + \frac{9}{10} A_8 \right) \]

This simplifies to:

\[ A_{12} = -\frac{1}{12} \tan^{11} x - \frac{11}{120} \tan^9 x + \frac{99}{120} A_8 \]

### Step 3: Combine \( A_{10} \) and \( A_{12} \)

Now we combine \( A_{10} \) and \( A_{12} \):

\[ A_{10} + A_{12} = \left( -\frac{1}{10} \tan^9 x + \frac{9}{10} A_8 \right) + \left( -\frac{1}{12} \tan^{11} x - \frac{11}{120} \tan^9 x + \frac{99}{120} A_8 \right) \]

Combining the terms gives:

\[ A_{10} + A_{12} = -\frac{1}{12} \tan^{11} x - \left( \frac{12}{120} + \frac{11}{120} \right) \tan^9 x + \left( \frac{9}{10} + \frac{99}{120} \right) A_8 \]

### Step 4: Simplify the expression

Now, we simplify the coefficients:

1. For \( \tan^9 x \):

\[ -\left( \frac{12 + 11}{120} \right) \tan^9 x = -\frac{23}{120} \tan^9 x \]

2. For \( A_8 \):

\[ \frac{9}{10} + \frac{99}{120} = \frac{108 + 99}{120} = \frac{207}{120} A_8 \]

### Step 5: Final expression

Thus, we have:

\[ A_{10} + A_{12} = -\frac{1}{12} \tan^{11} x - \frac{23}{120} \tan^9 x + \left( \frac{207}{120} A_8 \right) \]

### Step 6: Compare with given expression

We know that:

\[ A_{10} + A_{12} = \frac{\tan^m x}{m} + \lambda \]

Comparing the coefficients, we see that the term with \( \tan^{11} x \) corresponds to \( \frac{\tan^{11} x}{11} \). Thus, we conclude that:

\[ m = 11 \]

### Conclusion

The value of \( m \) is:

\[ \boxed{11} \]