Integration help!
integral of sec^5 x dx

- anonymous

Integration help!
integral of sec^5 x dx

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- katieb

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- anonymous

/barf im guessing u sub will help you here

- across

Integration by parts!

- across

Whoops, I misread.

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## More answers

- anonymous

yes integration by parts

- anonymous

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

- across

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

- anonymous

red formula will work if you have it

- anonymous

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

- anonymous

sec^2x = 1 + tan ^2

- anonymous

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

- across

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

- anonymous

this is wat i have so far:
\[\int\limits_{}^{}\sec^3 x (\sec^2 dx)\]
\[\int\limits_{}^{}\sec^3 x (1+\tan^2 x dx)\]
\[\int\limits_{}^{}(\sec^3 x +\sec^3 x \tan^2 x )dx\]

- anonymous

derive tan x = sec^2x

- anonymous

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

- anonymous

\[\int\limits_{}^{}(\sec^3 x) + \int\limits_{}^{}(\sec^3 x \tan^2 x) dx\]
can i use u sub after this step

- anonymous

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

- anonymous

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

- anonymous

or wat to do?

- anonymous

hang on...

- anonymous

im a little rusty

- anonymous

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

- anonymous

yea, it may have to be broken up into parts I cant get rid of one of the sec^x
check out: http://www.wolframalpha.com/input/?i=integrate+sec^5+x

- anonymous

this IS a reduction formula for powers of sec.

- anonymous

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

- anonymous

then again do u mind tellin me wat that is

- anonymous

or how to use it

- anonymous

yea, your instructor may not accept a reduction formula answer, it the quick way, but there is a trigonometric solution for this.

- anonymous

cool i will prob understand that if u explain

- anonymous

the trig solution i mean

- anonymous

there is an identity that will lead to an integral of u.
anywa here is a like with the reduction formula for secant, and others
http://www.sosmath.com/calculus/integration/moretrigpower/moretrigpower.html

- anonymous

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

- anonymous

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

- anonymous

a=1

- anonymous

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)

- anonymous

where do u go on from there

- anonymous

omg i havent done definite integrals!

- anonymous

no, I mistyped

- anonymous

∫sec^5dx=1/4sec^3xtanx+3/4∫sec^3xdx

- anonymous

no definite integrals here

- anonymous

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

- anonymous

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

- anonymous

im confused how u got the previous step thou

- anonymous

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

- anonymous

i mean teh 1/4 sec part im confused about

- anonymous

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

- anonymous

ok so now how did u get 3/4?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yea the trigonometric integrals take practice( obviously )

- anonymous

srry this is new to me so after taht step wat do u do..u still have a integral of sec^3 dx to deal with

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

wait but isnt is half secant

- anonymous

so how is teh integral jsut sec

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

1/4 sec^3 x tanx + 3/4∫sec^3xdx
then apply the formula again for ∫sec^3xdx
that answer will end with ∫sec xdx
integrate ∫secxdx, done!

- anonymous

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

- anonymous

SRY TAHTS SUPPOOSED TO be ln not tn

- anonymous

right, distribute the 3/4, I didnt work it all out but it looks right.
Those tables are in you textbook right, in the appendix?

- anonymous

let me check now

- anonymous

"table of integrals"

- anonymous

under wat heading shud i looj for te hreduction formulas

- anonymous

i see basic forms, forms invlving sqrt of (a^2+u^2, forms involving (a^2 - u^2),
forms involvung sqrt of (u^2 - a^2), forms involving a+bu, trig forms, inverse trig forms,
exp and log forms, hyperbolic forms

- anonymous

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

- anonymous

yeah i see them in the appenxix...jsut under wat heading of integrals will i fimd teh reduction formulas

- anonymous

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

- anonymous

that really what we were using, an integration formula, so you can integrate by parts or look up the formula.

- anonymous

there might be one for ∫sec^n (ax)dx

- anonymous

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

- anonymous

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

- anonymous

i guess tahts anoterh form of te one u showed me

- anonymous

right that one dont account for a coefficient on x.
so, yea, you seem to have it. Im out. ( gotta brush up on my integrals ;) )

- anonymous

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

- anonymous

peace

- anonymous

ii imma go study sum more

- anonymous

gnite

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