If I n = ∫ 0 π 4 tan n x d x , then 1 I 2 + 1 4 + 1 I 3 + I 5 + 1 I 4 + 1 6 =

Step-by-step solution

Given I n = ∫ 0 π 4 tan n x d x I n + I n + 2 = ∫ 0 π 4 tan n x d x + ∫ 0 π 4 tan n + 2 x d x = ∫ 0 π 4 tan n x sec 2 x d x = ∫ 0 1 t n d t where tan x = t = t n + 1 n + 1 0 1 = 1 n + 1 Now, 1 I 2 + 1 4 + 1 I 3 + I 5 + 1 I 4 + 1 6 = 3 + 4 + 5 = 12 = 1 I 11 + I 13