help_with_math
Integration help!
integral of sec^5 x dx
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Walleye
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/barf im guessing u sub will help you here
across
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Integration by parts!
across
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Whoops, I misread.
help_with_math
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yes integration by parts
Roachie
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(sec^2x)(sec^2)(sec x)
across
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I read it as sec^5(x)x for a moment there. Hmm, in this case, you have to use a reduction formula.
Walleye
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red formula will work if you have it
help_with_math
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well i havent learned taht so is tehre anoterh way to do it by parts
Roachie
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sec^2x = 1 + tan ^2
Roachie
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( 1 + tan ^2 x ) ( 1 + tan^2 x ) (sec x )
across
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I had to look it up\[\frac{\sin(x)\sec^{m-1}(x)}{m-1}+\frac{m+2}{m-1}\int\sec^{m-2}(x)dx\]
help_with_math
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this is wat i have so far:
\[\int\limits_{}^{}\sec^3 x (\sec^2 dx)\]
\[\int\limits_{}^{}\sec^3 x (1+\tan^2 x dx)\]
\[\int\limits_{}^{}(\sec^3 x +\sec^3 x \tan^2 x )dx\]
Roachie
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derive tan x = sec^2x
Roachie
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so do a u sub with u = tan x , du = sec^2 x dx
help_with_math
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\[\int\limits_{}^{}(\sec^3 x) + \int\limits_{}^{}(\sec^3 x \tan^2 x) dx\]
can i use u sub after this step
Roachie
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you will prolly end up with a ( 1 + u ) ^(to some power ) then it a simple integration
help_with_math
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can u show me the steps after the one i just did
help_with_math
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or wat to do?
Roachie
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hang on...
Roachie
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im a little rusty
help_with_math
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ughh i hate problems liek this!! XD...requires too much work
Roachie
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this IS a reduction formula for powers of sec.
help_with_math
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i did n idont understand it b.c it uses smthng i didnt leard which is teh reduction formula thing
help_with_math
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then again do u mind tellin me wat that is
help_with_math
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or how to use it
Roachie
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yea, your instructor may not accept a reduction formula answer, it the quick way, but there is a trigonometric solution for this.
help_with_math
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cool i will prob understand that if u explain
help_with_math
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the trig solution i mean
Roachie
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as a matter of fact I bet your textbook has a table of integrals for integrals of powers of secant
help_with_math
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what is the dffnce between sex^n (x) and sec^n (ax)
Roachie
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a=1
help_with_math
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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)
help_with_math
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where do u go on from there
help_with_math
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omg i havent done definite integrals!
Roachie
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no, I mistyped
Roachie
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∫sec^5dx=1/4sec^3xtanx+3/4∫sec^3xdx
Roachie
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no definite integrals here
help_with_math
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hmm how did u get 1/4 sec ...etc
Roachie
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then apply the reduction formula ( or integrate ) \[\int\limits \sec ^ 3 x dx\]
help_with_math
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im confused how u got the previous step thou
Roachie
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its from the reduction formula, it ends with an integral of ∫sec^3xdx
help_with_math
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i mean teh 1/4 sec part im confused about
Roachie
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in the reduction formula it 1/(a(n-1)
n=5, the power of secant
a = 1
help_with_math
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ok so now how did u get 3/4?
Roachie
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from the reduction formula: (n-2)/(n-1)
again n = 5, the power of the secant
Roachie
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you instructor will be either impressed or put off that you are using the integral tables. :)
help_with_math
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lol...he will b impressed hopefully...i have a test 2mm n im trying to udnerstand probs idk
Roachie
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yea the trigonometric integrals take practice( obviously )
help_with_math
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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
Roachie
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right! so integrate the sec^x or use the formula again, but this time with n = 3, see?
Roachie
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sec^x is short, but it indefinite integral has many parts.
help_with_math
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so after using n=3 do u get 1/2 sec x dx
Roachie
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right ( n-2)/(n-1)
this time the formula will end with \[\int\limits \sec x dx\]
help_with_math
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wait but isnt is half secant
help_with_math
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so how is teh integral jsut sec
Roachie
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no the formula will end ith ∫secxdx, integrate that and you are done.
Roachie
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\[\int\limits \sec x dx = \ln ( \sec x +tanx ) + C\]
help_with_math
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so teh final answer is: 1/4 sec^3 x tanx + 3/4 ...idk teh rest
Roachie
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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!
help_with_math
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so the final asnwer is:
1/4 sec^3 x tanx + 3/8 sec x tanx + tn (secx+tanx) + C
help_with_math
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SRY TAHTS SUPPOOSED TO be ln not tn
Roachie
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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?
help_with_math
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let me check now
Roachie
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"table of integrals"
help_with_math
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under wat heading shud i looj for te hreduction formulas
help_with_math
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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
Roachie
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Tables of integrals, Lists of Integrals usually in the appendix, could be on the inside covers.
help_with_math
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yeah i see them in the appenxix...jsut under wat heading of integrals will i fimd teh reduction formulas
help_with_math
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or wher u jsut letting me know taht tehre was a table of integrtals
Roachie
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that really what we were using, an integration formula, so you can integrate by parts or look up the formula.
Roachie
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there might be one for ∫sec^n (ax)dx
help_with_math
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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
help_with_math
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i guess tahts anoterh form of te one u showed me
Roachie
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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 ;) )
help_with_math
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THANK U SOOOOO MUCH!!! PLUS THE WEBSITE LOOKS SUPER HELPFUL =)
Roachie
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peace
help_with_math
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ii imma go study sum more
help_with_math
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gnite