Question

If $\int_0^{\pi/2}\tan^{14}(x/2)dx = 2\left[\sum_{n=1}^7 f(n) - \frac{\pi}{4}\right]$, then $f(n)=$

• A. $\frac{(-1)^n}{n-1}$
• B. $\frac{(-1)^n}{2n+1}$
• C. $\frac{(-1)^{n+1}}{2n-1}$
• D. $\frac{(-1)^{n+1}}{n+1}$

Hint:

The reduction formula for the definite integral of tangent, $J_m = \int_0^{\pi/4} \tan^m(x) dx$, is $J_m + J_{m-2} = \frac{1}{m-1}$. This is a very useful formula for evaluating integrals of powers of tangent over this specific interval.

Answer 1

The Correct Option is C

Step 1: Set up the integral and use a substitution.
Let $I = \int_0^{\pi/2} \tan^{14}(x/2)dx$. Let $u = x/2$, so $dx = 2du$. When $x=0, u=0$. When $x=\pi/2, u=\pi/4$. \[ I = \int_0^{\pi/4} \tan^{14}(u) (2du) = 2 \int_0^{\pi/4} \tan^{14}(u) du. \]

Step 2: Use the reduction formula for the integral of $\tan^m(u)$.
Let $J_m = \int_0^{\pi/4} \tan^m(u) du$. The reduction formula is $J_m = \frac{1}{m-1} - J_{m-2}$ for $m \ge 2$. We need to calculate $I = 2J_{14}$. $J_0 = \int_0^{\pi/4} \tan^0(u) du = \int_0^{\pi/4} 1 du = [u]_0^{\pi/4} = \pi/4$.

Step 3: Apply the reduction formula repeatedly.
$J_{14} = \frac{1}{13} - J_{12}$ $J_{12} = \frac{1}{11} - J_{10}$ ... $J_2 = \frac{1}{1} - J_0 = 1 - \frac{\pi}{4}$. Now, let's express $J_{14}$ as a sum: $J_{14} = \frac{1}{13} - J_{12} = \frac{1}{13} - (\frac{1}{11} - J_{10}) = \frac{1}{13} - \frac{1}{11} + J_{10}$. Continuing this process: \[ J_{14} = \frac{1}{13} - \frac{1}{11} + \frac{1}{9} - \frac{1}{7} + \frac{1}{5} - \frac{1}{3} + J_2. \] \[ J_{14} = \frac{1}{13} - \frac{1}{11} + \frac{1}{9} - \frac{1}{7} + \frac{1}{5} - \frac{1}{3} + (1 - \frac{\pi}{4}). \] Rearranging the terms: \[ J_{14} = \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13}\right) - \frac{\pi}{4}. \]

Step 4: Compare with the given expression.
The original equation is $I = 2J_{14} = 2\left[\sum_{n=1}^7 f(n) - \frac{\pi}{4}\right]$. This implies that $J_{14} = \sum_{n=1}^7 f(n) - \frac{\pi}{4}$. Comparing this with our result for $J_{14}$: \[ \sum_{n=1}^7 f(n) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13}. \]

Step 5: Identify the general term $f(n)$.
The sum is an alternating series of reciprocals of odd numbers. The $n^{th}$ term of the sequence $1, 3, 5, \dots$ is $2n-1$. The sign of the $n^{th}$ term is positive if $n$ is odd and negative if $n$ is even. This corresponds to the factor $(-1)^{n+1}$. So, the general term is $f(n) = \frac{(-1)^{n+1}}{2n-1}$. This matches option (C).