If `I_(n)=int_(0)^(pi//4)tan^(n)x dx,` then `7(I_(6)+I_(8))=`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \) and find \( 7(I_6 + I_8) \), we will follow these steps:

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

Using the reduction formula for integrals of \( \tan^n x \):
\[ I_n = \frac{n-1}{n} I_{n-2} \]
We will first find \( I_6 \) and \( I_8 \).

### Step 2: Calculate \( I_6 \)

Using the reduction formula:
\[ I_6 = \frac{5}{6} I_4 \]

### Step 3: Calculate \( I_4 \)

Using the reduction formula again:
\[ I_4 = \frac{3}{4} I_2 \]

### Step 4: Calculate \( I_2 \)

Using the reduction formula once more:
\[ I_2 = \frac{1}{2} I_0 \]

### Step 5: Calculate \( I_0 \)

We know:
\[ I_0 = \int_0^{\frac{\pi}{4}} \tan^0 x \, dx = \int_0^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4} \]

### Step 6: Substitute back to find \( I_2 \)

Now substituting back:
\[ I_2 = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8} \]

### Step 7: Substitute to find \( I_4 \)

Now substituting \( I_2 \) back:
\[ I_4 = \frac{3}{4} \cdot \frac{\pi}{8} = \frac{3\pi}{32} \]

### Step 8: Substitute to find \( I_6 \)

Now substituting \( I_4 \) back:
\[ I_6 = \frac{5}{6} \cdot \frac{3\pi}{32} = \frac{15\pi}{192} = \frac{5\pi}{64} \]

### Step 9: Calculate \( I_8 \)

Using the reduction formula again:
\[ I_8 = \frac{7}{8} I_6 = \frac{7}{8} \cdot \frac{5\pi}{64} = \frac{35\pi}{512} \]

### Step 10: Calculate \( 7(I_6 + I_8) \)

Now we can calculate:
\[ I_6 + I_8 = \frac{5\pi}{64} + \frac{35\pi}{512} \]

To add these fractions, we need a common denominator:
\[ \frac{5\pi}{64} = \frac{40\pi}{512} \]
Thus,
\[ I_6 + I_8 = \frac{40\pi}{512} + \frac{35\pi}{512} = \frac{75\pi}{512} \]

Finally, calculate \( 7(I_6 + I_8) \):
\[ 7(I_6 + I_8) = 7 \cdot \frac{75\pi}{512} = \frac{525\pi}{512} \]

### Final Answer

Thus, the final answer is:
\[ 7(I_6 + I_8) = \frac{525\pi}{512} \]