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Help!! Integration Question (1 Viewer)

heybashme

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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
 

conics2008

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ok this is what you do

write (tanx)^2n as (tanx)^2n-2 . tanx^2

therefore tanx^2 = sec^2x-1

therefore when you expand it, you will get tanx^2n-2 = I n-2

and tan^n-2 sec^2

take I n-2 to the other side. and then intergrate tan^n-2sec^2 using u=tan

when you do that you will get that.

hope you understood
 

ronnknee

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conics2008 said:
take I n-2 to the other side. and then intergrate tan^n-2sec^2 using u=tan
When x = pi/2, u = undefined
 

3unitz

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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)
 
Last edited:

conics2008

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ronnknee said:
When x = pi/2, u = undefined
its pi/4 .. they wont give you limits between pi/2, because the top of tan^2n-2 has to equal to 1.. therefore it must be pi/4...

i never said anything about pi/2

the guy put it there. i showed him how to do it.:read:
 

heybashme

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thanx for everyones help but yeh its supposed to be pi/4 not pi/2 my mistake
thanx 4 the help guys btw
 

conics2008

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ronkkee i know, i just ignored it, because i know tan90 dont exsist, soo yeh xD.... these types of question you just need to know your identites.
 

carmey

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idk if you still need help on this questions but here it is..

[edit] PS i kinda assume u mean pi/4
 

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