Question

If $I_{n}=\int_{0}^{\pi/4}\tan^{n}x~dx,$ then $I_{13}+I_{11}=$

• A. $\frac{1}{13}$
• B. $\frac{1}{12}$
• C. $\frac{1}{10}$
• D. $\frac{1}{11}$

Hint:

Use integration by parts or a suitable substitution to derive the reduction formula for I_n.

Answer 1

The Correct Option is B

Step 1: Find a reduction formula for I_n.
I_n = $\int_{0}^{\pi/4} \tan^n x \, dx$ = $\int_{0}^{\pi/4} \tan^{n-2} x \tan^2 x \, dx$
I_n = $\int_{0}^{\pi/4} \tan^{n-2} x (\sec^2 x - 1) \, dx$
I_n = $\int_{0}^{\pi/4} \tan^{n-2} x \sec^2 x \, dx - \int_{0}^{\pi/4} \tan^{n-2} x \, dx$
I_n = $\int_{0}^{\pi/4} \tan^{n-2} x \sec^2 x \, dx - I_{n-2}$
Let u = tan x, then du = sec² x dx.
When x = 0, u = 0. When x = π/4, u = 1.
I_n = $\int_{0}^{1} u^{n-2} \, du - I_{n-2}$
I_n = $\left[\frac{u^{n-1}}{n-1}\right]_{0}^{1} - I_{n-2}$
I_n = $\frac{1}{n-1} - I_{n-2}$

Step 2: Use the reduction formula to find I₁₃ + I₁₁.
Using the reduction formula, we have:
I₁₃ = $\frac{1}{13-1} - I_{11}$ = $\frac{1}{12} - I_{11}$
I₁₃ + I₁₁ = $\frac{1}{12}$
Therefore, I₁₃ + I₁₁ = $\frac{1}{12}$.