wtffffffffff (1 Viewer)

[kunny funt]

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]

it involves careful reading of the question paper

 

[nit]

Originally posted by Slide Rule

OK, tywebb=nanahcub=buchanan=Derek!

Well that proves it eh?

 

[mojako]

i say what u know isn't important.

i say what u believe is important.

umm.. I have trouble distinguishing between the two....

 

[Estel]

Who cares?
Does it really matter?

Just plunder the solutions and be happy.

 

[SipSip]

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

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

 

[mojako]

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