IF `I_n=int_0^(pi//4) tan^n x dx` then what is `I_n+I_(n+2)` equal to

Options

A. A. `1/n`
B. B. `1/((n+1))`
C. C. `n/((n-1))`
D. D. `1/((n-2))`

Answer

To solve the problem, we need to evaluate \( I_n + I_{n+2} \) where \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \).

### Step 1: Write down the expressions for \( I_n \) and \( I_{n+2} \)
We have:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]
\[ I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^{n+2} x \, dx \]

### Step 2: Combine the integrals
Now, we want to find \( I_n + I_{n+2} \):
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \, dx + \int_0^{\frac{\pi}{4}} \tan^{n+2} x \, dx \]
This can be combined into a single integral:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \left( \tan^n x + \tan^{n+2} x \right) \, dx \]

### Step 3: Factor out \( \tan^n x \)
We can factor out \( \tan^n x \):
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \left( 1 + \tan^2 x \right) \, dx \]

### Step 4: Use the identity \( 1 + \tan^2 x = \sec^2 x \)
Using the trigonometric identity \( 1 + \tan^2 x = \sec^2 x \), we can rewrite the integral:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \sec^2 x \, dx \]

### Step 5: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \). The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \)

Thus, we have:
\[ I_n + I_{n+2} = \int_0^1 t^n \, dt \]

### Step 6: Evaluate the integral
Now we can evaluate the integral:
\[ \int_0^1 t^n \, dt = \left[ \frac{t^{n+1}}{n+1} \right]_0^1 = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} \]

### Final Result
Thus, we find:
\[ I_n + I_{n+2} = \frac{1}{n+1} \]