heybashme said:
plz if some1 could help ive literally spent about an hour and a half trying to solve it. thanks in advance
If In= Л/2[integral sign]0 (tan x)^2n dx
show that In + In-2 = 1/(2n-1)
i bet theres an easy method but im jus blind
hmm, not sure your question is correct, im going to assume pi/4
ignoring limits for now:
I (tan x)^2n dx
= I (tan x)^(2n - 2) (tan x)^2 dx
= I (tan x)^(2n - 2) [(sec x)^2 - 1] dx
= I (tan x)^(2n - 2)(sec x)^2 dx - I (tan x)^(2n - 2) dx
In + I(n-2) = I (tan x)^(2n - 2)(sec x)^2 dx
we need to find:
{limits pi/4 and 0} I (tan x)^(2n - 2)(sec x)^2 dx
[(tan x)^(2n - 1) / (2n - 1)] {0 -> pi/4}
= 1/(2n - 1)
In + I(n-2) = 1/(2n - 1)