MedVision ad

integration question. (1 Viewer)

bboyelement

Member
Joined
May 3, 2005
Messages
242
Gender
Male
HSC
2006
integrate (sec)^3

i got solution to it but i dont understand how they got it ... i just need explanation.
thanks
 

felixcthecat

Member
Joined
Mar 4, 2005
Messages
308
Gender
Male
HSC
2006
typed out so nicely :D
and a second 'yup' its correct~~

riviet.. is there another method? this would be the only method i could think of
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
Not really... this way seems like the most efficient, in my opinion.
 

bboyelement

Member
Joined
May 3, 2005
Messages
242
Gender
Male
HSC
2006
actually theres another way ... but your way is much easier i think... thanks
 

Wackedupwacko

Member
Joined
Nov 29, 2005
Messages
141
Location
Sydney
Gender
Male
HSC
2006
there is another way... mass reduction but thats not really recommended since its not a fun thing.
S -> integral sign
In = S sec^n x dx
= S (sec^2 x)(sec^(n-2) x) dx
= tan x sec^(n-2) x - (n-2)S tan^2 x sec^(n-2) x dx
= tan x sec^(n-2) x - (n-2)S (sec^2 x - 1) (sec^(n-2) x) dx
= tan x sec^(n-2) x - (n-2)S sec^n x - sec^(n-2) x dx
= tan x sec^(n-2) x - (n-2)(In - I(n-2))
(n-2)In +In = tanx sec^(n-2) x + (n-2)I(n-2)
In = (tanx sec^(n-2) x + (n-2)I(n-2) )/ (n-1)

in which u just plug in numbers to that formula and ull get urself an answer... which in this case n = 3 so

I3 = tanxsecx + I1 / 2
I1 = ln| tanx + secx| + C thus:

I3 = (tanxsecx + ln|tanx + secx| )/ 2 + c

of course this is a long way to do things if u dunno the reduction formula but... its just another way

the thing with this is u cna use it for any power of sec so thats useful =]
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top