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/barf im guessing u sub will help you here

Integration by parts!

Whoops, I misread.

yes integration by parts

(sec^2x)(sec^2)(sec x)

I read it as sec^5(x)x for a moment there. Hmm, in this case, you have to use a reduction formula.

red formula will work if you have it

well i havent learned taht so is tehre anoterh way to do it by parts

sec^2x = 1 + tan ^2

( 1 + tan ^2 x ) ( 1 + tan^2 x ) (sec x )

I had to look it up\[\frac{\sin(x)\sec^{m-1}(x)}{m-1}+\frac{m+2}{m-1}\int\sec^{m-2}(x)dx\]

derive tan x = sec^2x

so do a u sub with u = tan x , du = sec^2 x dx

you will prolly end up with a ( 1 + u ) ^(to some power ) then it a simple integration

can u show me the steps after the one i just did

or wat to do?

hang on...

im a little rusty

ughh i hate problems liek this!! XD...requires too much work

this IS a reduction formula for powers of sec.

i did n idont understand it b.c it uses smthng i didnt leard which is teh reduction formula thing

then again do u mind tellin me wat that is

or how to use it

cool i will prob understand that if u explain

the trig solution i mean

as a matter of fact I bet your textbook has a table of integrals for integrals of powers of secant

what is the dffnce between sex^n (x) and sec^n (ax)

a=1

i wud use the first formula right?
sec^5 (x) = sec ^5-2 (x) sec^2(x) = sec^3 (tan^2 x * sec^2 x -1)

where do u go on from there

omg i havent done definite integrals!

no, I mistyped

∫sec^5dx=1/4sec^3xtanx+3/4∫sec^3xdx

no definite integrals here

hmm how did u get 1/4 sec ...etc

then apply the reduction formula ( or integrate ) \[\int\limits \sec ^ 3 x dx\]

im confused how u got the previous step thou

its from the reduction formula, it ends with an integral of ∫sec^3xdx

i mean teh 1/4 sec part im confused about

in the reduction formula it 1/(a(n-1)
n=5, the power of secant
a = 1

ok so now how did u get 3/4?

from the reduction formula: (n-2)/(n-1)
again n = 5, the power of the secant

you instructor will be either impressed or put off that you are using the integral tables. :)

lol...he will b impressed hopefully...i have a test 2mm n im trying to udnerstand probs idk

yea the trigonometric integrals take practice( obviously )

right! so integrate the sec^x or use the formula again, but this time with n = 3, see?

sec^x is short, but it indefinite integral has many parts.

so after using n=3 do u get 1/2 sec x dx

right ( n-2)/(n-1)
this time the formula will end with \[\int\limits \sec x dx\]

wait but isnt is half secant

so how is teh integral jsut sec

no the formula will end ith ∫secxdx, integrate that and you are done.

\[\int\limits \sec x dx = \ln ( \sec x +tanx ) + C\]

so teh final answer is: 1/4 sec^3 x tanx + 3/4 ...idk teh rest

so the final asnwer is:
1/4 sec^3 x tanx + 3/8 sec x tanx + tn (secx+tanx) + C

SRY TAHTS SUPPOOSED TO be ln not tn

let me check now

"table of integrals"

under wat heading shud i looj for te hreduction formulas

Tables of integrals, Lists of Integrals usually in the appendix, could be on the inside covers.

or wher u jsut letting me know taht tehre was a table of integrtals

there might be one for ∫sec^n (ax)dx

might not
http://integral-table.com/

ths is wat is in my bk...sec^n du = 1/n tan u sec^n-2 + (n-2/n-1) *integral of sec^n-2 u du

i guess tahts anoterh form of te one u showed me

THANK U SOOOOO MUCH!!! PLUS THE WEBSITE LOOKS SUPER HELPFUL =)

peace

ii imma go study sum more

gnite