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kunny funt

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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

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it involves careful reading of the question paper
 

mojako

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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

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Who cares?
Does it really matter?

Just plunder the solutions and be happy.
 

SipSip

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

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

mojako

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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
 
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