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integration by parts qu (2 Viewers)

asb_92

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hey guys

was wondering how do u integrate:

sec³x (sec cubed x)

using integration by parts?:confused:

thanks heaps =]
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annabackwards

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Separate (sec x)^3 into sec x ( (sec x)^2) = sec x (1 + (tan x)^2) = sec x + sec x (tan x)^2

Then just integrate sec x + sec x (tan x)^2 normally :)
 

Dumbledore

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split it into S[sec(x)sec^2(x)dx]
then S[sec(x)d(tan(x))]
S[sec^3(x)]= sec(x)tan(x)-S[sec(x)tan^2(x)]
----------S[sec(x)tan^2(x)] = S(sec(x)*[sec^2(x)-1])=S(sec^3(x))-S[sec(x)
= sec(x)tan(x) - S(sec^3(x)+S[sec(x)] take S[sec^3(x) to other side
2S[sec^3(x)] = sec(x)tan(x) + ln[sec(x)+tan(x)]
S[sec^3(x)] = 1/2sec(x)tan(x) + 1/2ln[sec(x)+tan(x)]
 
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asb_92

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ah right
wow
thanks heaps guys :D
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GUSSSSSSSSSSSSS

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Separate (sec x)^3 into sec x ( (sec x)^2) = sec x (1 + (tan x)^2) = sec x + sec x (tan x)^2

Then just integrate sec x + sec x (tan x)^2 normally :)
that part wud actually need to be (secx)^2 . (tanx)^2 as the derivative of tanx is (secx)^2 =[[[
 

annabackwards

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ye ye thats perfect answer ...... but is that wat u were saying in ur first post ??? =S i cant tell lol too tired =P damn english !!
Aha, i'm not sure myself. I was probably only awake then from all the coffee i had so i probably said something retarded =="
 

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