kunny funt said:

somebody explain to me how u get : I[n] + I[n-2] = 1/n+1 for the integral of tan^n [x] between 0 and pi/4

I[n] = integral (tan^n-2 [x])(sec^2 [x] - 1) coz t^2 = sec^2 -1 etc

= integral (tan^n-2 [x])(sec^2 [x]) - I[n-2]

I[n] + I[n-2] = (1/n-1)(tan^n-1 [x]) for 0 to pi/4

= 1/n-1 - 0

= 1/n-1

i srsly dont get how its meant to be a +

ok, since nobody has actually "answered" this Q on the forum, and since tywebb's solution may disappear at any time,

your result is true.

I[n] + I[n-2] = 1/(n-1)

now, replace "n" with "n+2"

and you get the required result.

To SipSip:

there is an axiom we learnt that integra=integral