# wtffffffffff (1 Viewer)

#### kunny funt

##### Large Member
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 +

#### Rorix

##### Active Member
it involves careful reading of the question paper

#### nit

##### Member
Originally posted by Slide Rule

OK, tywebb=nanahcub=buchanan=Derek!
Well that proves it eh? #### mojako

##### Active Member
tywebb said:
i say what u know isn't important.

i say what u believe is important.
umm.. I have trouble distinguishing between the two....

#### Estel

##### Tutor
Who cares?
Does it really matter?

Just plunder the solutions and be happy.

#### SipSip

##### Member
wtf is an integral.....i know an integra....

shit man!! i did 4U last year and i've forgotten everything...

#### mojako

##### Active Member
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,