If `I_(n)=overset(pi//4)underset(0)int tan ^(n) x dx, lim_(n to oo) n(I_(n+1)+I_(n-1))` equals

Options

A. `(1)/(2)`
B. 1
C. `oo`
D. zero

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} n(I_{n+1} + I_{n-1}) \]

where

\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx. \]

### Step 1: Express \( I_n \) in terms of \( I_{n-2} \)

Using the identity for \( \tan^n x \):

\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \tan^2 x \, dx \]

We can rewrite \( \tan^2 x \) as \( \sec^2 x - 1 \):

\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot (\sec^2 x - 1) \, dx \]

This expands to:

\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]

The first integral can be rewritten as:

\[ I_n = I_{n-2} + I_{n-2} \]

Thus, we have:

\[ I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx \]

### Step 2: Change of Variables

Now, we use the substitution \( t = \tan x \), hence \( dt = \sec^2 x \, dx \).

The limits change as follows:

- When \( x = 0 \), \( t = 0 \)
- When \( x = \frac{\pi}{4} \), \( t = 1 \)

Thus, we have:

\[ I_n + I_{n-2} = \int_0^1 t^{n-2} \, dt \]

### Step 3: Evaluate the Integral

The integral evaluates to:

\[ \int_0^1 t^{n-2} \, dt = \left[ \frac{t^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1} \]

So we have:

\[ I_n + I_{n-2} = \frac{1}{n-1} \]

### Step 4: Relate \( I_{n+1} \) and \( I_{n-1} \)

By a similar argument, we can write:

\[ I_{n+1} + I_{n-1} = \frac{1}{n} \]

### Step 5: Combine the Results

Now we have:

\[ I_n + I_{n-2} = \frac{1}{n-1} \]
\[ I_{n+1} + I_{n-1} = \frac{1}{n} \]

### Step 6: Find the Limit

Now, we need to find:

\[ \lim_{n \to \infty} n(I_{n+1} + I_{n-1}) = \lim_{n \to \infty} n \cdot \frac{1}{n} = 1 \]

### Conclusion

Thus, the final answer is:

\[ \lim_{n \to \infty} n(I_{n+1} + I_{n-1}) = 1 \]