Question

If \(I_n=^{\frac{\pi}{4}}_0\int\tan^n x\ dx\) where n is positive integer then I₁₀ + I₈ is equal to

• A. 9
• B. \(\frac{1}{7}\)
• C. \(\frac{1}{8}\)
• D. \(\frac{1}{9}\)

Answer 1

The Correct Option is D

We are given the definition of an integral \( I_n \) as:

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

where \( n \) is a positive integer. We need to find the value of \( I_{10} + I_8 \).

Step 1: Establish a recurrence relation for \( I_n \).

Consider the sum \( I_n + I_{n-2} \) for \( n \ge 2 \):

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

Combine the integrals:

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

Factor out \( \tan^{n-2}(x) \):

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

Use the identity \( \tan^2(x) + 1 = \sec^2(x) \):

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

Step 2: Evaluate the integral using substitution.

Let \( u = \tan(x) \). Then \( du = \sec^2(x) \, dx \).

Change the limits of integration:

• When \( x = 0 \), \( u = \tan(0) = 0 \).
• When \( x = \pi/4 \), \( u = \tan(\pi/4) = 1 \).

Substitute into the integral:

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

Evaluate the integral:

\[ I_n + I_{n-2} = \left[ \frac{u^{n-2+1}}{n-2+1} \right]_0^1 = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 \] \[ I_n + I_{n-2} = \frac{1^{n-1}}{n-1} - \frac{0^{n-1}}{n-1} \]

Since \( n \ge 2 \), \( n-1 \ge 1 \), so \( 0^{n-1} = 0 \).

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

This is the recurrence relation, valid for \( n \ge 2 \).

Step 3: Apply the recurrence relation to find \( I_{10} + I_8 \).

We need to find \( I_{10} + I_8 \). This matches the form \( I_n + I_{n-2} \) with \( n = 10 \).

Using the relation \( I_n + I_{n-2} = \frac{1}{n-1} \) with \( n = 10 \):

\[ I_{10} + I_{10-2} = \frac{1}{10-1} \] \[ I_{10} + I_8 = \frac{1}{9} \]

So, the correct answer is (D): ${\frac{1}{9}}$

Answer 2

Given:
$ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx $ We are asked to find:
$ I_{10} + I_8 $ 

Step 1: Use the reduction formula
The standard reduction formula is:
$ I_n = \frac{1}{n - 1} - I_{n - 2} $ 

Step 2: Apply the formula recursively
1. $ I_0 = \int_0^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4} $ 2. $ I_2 = \frac{1}{1} - I_0 = 1 - \frac{\pi}{4} $ 3. $ I_4 = \frac{1}{3} - I_2 = \frac{1}{3} - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2}{3} $ 4. $ I_6 = \frac{1}{5} - I_4 = \frac{1}{5} - \left(\frac{\pi}{4} - \frac{2}{3}\right) = \frac{13}{15} - \frac{\pi}{4} $ 5. $ I_8 = \frac{1}{7} - I_6 = \frac{1}{7} - \left(\frac{13}{15} - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{76}{105} $ 6. $ I_{10} = \frac{1}{9} - I_8 = \frac{1}{9} - \left(\frac{\pi}{4} - \frac{76}{105}\right) = \frac{181}{315} - \frac{\pi}{4} $ 

Step 3: Add $I_{10}$ and $I_8$
$ I_{10} + I_8 = \left( \frac{1}{9} - I_8 \right) + I_8 = \frac{1}{9} $ 

Final Answer:
$ \frac{1}{9} $