If `I_(n) = int _(0)^(pi//4) tan^(n) theta " " d theta, "then "I_(8)+I_(6)=`

Options

A. `1/4`
B. `1/5`
C. `1/6`
D. `1/7`

Answer

To solve the problem, we need to find the value of \( I_8 + I_6 \), where

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

### Step 1: Write the expression for \( I_8 + I_6 \)

We start by expressing \( I_8 + I_6 \):

\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^8 \theta \, d\theta + \int_0^{\frac{\pi}{4}} \tan^6 \theta \, d\theta. \]

### Step 2: Factor out \( I_6 \)

Notice that we can factor out \( I_6 \):

\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^6 \theta \, d\theta + \int_0^{\frac{\pi}{4}} \tan^6 \theta \cdot \tan^2 \theta \, d\theta. \]

This can be rewritten as:

\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^6 \theta \left( 1 + \tan^2 \theta \right) d\theta. \]

### Step 3: Use the identity \( 1 + \tan^2 \theta = \sec^2 \theta \)

Using the trigonometric identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify the integral:

\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^6 \theta \sec^2 \theta \, d\theta. \]

### Step 4: Substitute \( t = \tan \theta \)

Now we will use the substitution \( t = \tan \theta \), which gives us \( dt = \sec^2 \theta \, d\theta \). The limits change as follows:

- When \( \theta = 0 \), \( t = \tan(0) = 0 \).
- When \( \theta = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \).

Thus, we have:

\[ I_8 + I_6 = \int_0^1 t^6 \, dt. \]

### Step 5: Evaluate the integral

Now we can evaluate the integral:

\[ \int_0^1 t^6 \, dt = \left[ \frac{t^7}{7} \right]_0^1 = \frac{1^7}{7} - \frac{0^7}{7} = \frac{1}{7}. \]

### Conclusion

Thus, we find that:

\[ I_8 + I_6 = \frac{1}{7}. \]