Question

If \[ I_n=\int_{0}^{\pi/4}\tan^n x\,dx, \] then \[ \frac{1}{I_2+I_4}+\frac{1}{I_3+I_5}+\frac{1}{I_4+I_6} \] is equal to

• A. \(\dfrac{1}{I_9+I_{11}}\)
• B. \(\dfrac{1}{I_{10}+I_{12}}\)
• C. \(\dfrac{1}{I_{12}+I_{14}}\)
• D. \(\dfrac{1}{I_{11}+I_{13}}\)

Hint:

For \(I_n=\int_0^{\pi/4}\tan^n x\,dx\), the expression \(I_n+I_{n+2}\) simplifies easily because \(\tan^n x+\tan^{n+2}x=\tan^n x\sec^2x\).

Answer 1

The Correct Option is D

Step 1: Use the definition of \(I_n\).
Given, \[ I_n=\int_{0}^{\pi/4}\tan^n x\,dx \] Now, \[ I_n+I_{n+2} = \int_{0}^{\pi/4}\tan^n x\,dx+ \int_{0}^{\pi/4}\tan^{n+2}x\,dx \] \[ = \int_{0}^{\pi/4}\left(\tan^n x+\tan^{n+2}x\right)\,dx \] \[ = \int_{0}^{\pi/4}\tan^n x\left(1+\tan^2x\right)\,dx \] Since, \[ 1+\tan^2x=\sec^2x, \] we get \[ I_n+I_{n+2} = \int_{0}^{\pi/4}\tan^n x\sec^2x\,dx \]

Step 2: Evaluate \(I_n+I_{n+2}\).
Let \[ t=\tan x \] Then, \[ dt=\sec^2x\,dx \] When \[ x=0,\qquad t=0 \] and when \[ x=\frac{\pi}{4},\qquad t=1 \] Therefore, \[ I_n+I_{n+2} = \int_0^1 t^n\,dt \] \[ = \left[\frac{t^{n+1}}{n+1}\right]_0^1 \] \[ = \frac{1}{n+1} \] Thus, \[ I_n+I_{n+2}=\frac{1}{n+1} \]

Step 3: Use the result for each term.
For \(n=2\), \[ I_2+I_4=\frac{1}{3} \] Hence, \[ \frac{1}{I_2+I_4}=3 \] For \(n=3\), \[ I_3+I_5=\frac{1}{4} \] Hence, \[ \frac{1}{I_3+I_5}=4 \] For \(n=4\), \[ I_4+I_6=\frac{1}{5} \] Hence, \[ \frac{1}{I_4+I_6}=5 \] Therefore, \[ \frac{1}{I_2+I_4}+\frac{1}{I_3+I_5}+\frac{1}{I_4+I_6} = 3+4+5 \] \[ =12 \]

Step 4: Match the result with the options.
We need an option of the form \[ \frac{1}{I_n+I_{n+2}} \] Using \[ I_n+I_{n+2}=\frac{1}{n+1}, \] we get \[ \frac{1}{I_n+I_{n+2}}=n+1 \] Since the value is \(12\), \[ n+1=12 \] \[ n=11 \] Thus, \[ 12=\frac{1}{I_{11}+I_{13}} \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{1}{I_{11}+I_{13}}} \]