If `rArr I_(n)=int_(0)^(pi//4) tan ^(n)x dx`, then for any positive integer, n, the value of `n(I_(n+1)+I_(n-1))` is,

Options

A. 1
B. 2
C. `pi//4`
D. `pi`

Explanation

To solve the problem, we start with the integral defined as:

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

We need to find the value of \( n(I_{n+1} + I_{n-1}) \).

### Step 1: Express \( I_{n+1} \)

We can express \( I_{n+1} \) as follows:

\[ I_{n+1} = \int_0^{\frac{\pi}{4}} \tan^{n+1} x \, dx = \int_0^{\frac{\pi}{4}} \tan^n x \tan x \, dx \]

Using the identity \( \tan^2 x = \sec^2 x - 1 \), we can rewrite \( \tan^{n+1} x \):

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

This can be split into two integrals:

\[ I_{n+1} = \int_0^{\frac{\pi}{4}} \tan^n x \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]

Thus, we have:

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

### Step 2: Express \( I_{n-1} \)

Now, we express \( I_{n-1} \):

\[ I_{n-1} = \int_0^{\frac{\pi}{4}} \tan^{n-1} x \, dx \]

### Step 3: Combine \( I_{n+1} \) and \( I_{n-1} \)

Now, we combine \( I_{n+1} \) and \( I_{n-1} \):

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

This simplifies to:

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

### Step 4: Change of Variables

Next, we perform a change of variables. Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{4} \) which corresponds to \( t = 0 \) to \( t = 1 \).

Thus, we have:

\[ \int_0^{\frac{\pi}{4}} \tan^n x \sec^2 x \, dx = \int_0^1 t^n \, dt \]

Calculating this integral gives:

\[ \int_0^1 t^n \, dt = \frac{1}{n+1} \]

### Step 5: Substitute Back

Now substituting back, we have:

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

### Step 6: Final Calculation

Now we multiply the entire expression by \( n \):

\[ n(I_{n+1} + I_{n-1}) = n \left( \frac{1}{n+1} - I_n + I_{n-1} \right) \]

Since \( I_n \) and \( I_{n-1} \) terms will cancel out, we find:

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

### Final Answer

Thus, the value of \( n(I_{n+1} + I_{n-1}) \) is:

\[ \boxed{1} \]