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farukkBest ResponseYou've already chosen the best response.0
\[\int\limits \tan^3 x secx dx\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ \tan^3 x \sec x = \tan x \sec x ( \sec^2 x  1)\] seems like it will get the same
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
note that d(sec x) =sec x tan x dx
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
let u = sec x then proceed. du= sec x tan x dx so it all becomes \(\int (u^2  1) du\)
 one year ago

farukkBest ResponseYou've already chosen the best response.0
dear are you solving integral by parts?
 one year ago

farukkBest ResponseYou've already chosen the best response.0
it will become like this i guess u^3/3  u +c
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
\( \int (\sec^21) \tan x \sec x dx\) =\( \int (\sec^21) d(\sec x)\) =\( \int (u^21) du\), where u=\sec x\) by substitution.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
eh? \(\int u^n du= \frac{u^{n+1}}{n+1} +C\)
 one year ago

farukkBest ResponseYou've already chosen the best response.0
after applying we ll get as i above wrote ?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
you'll get the answer....from integrating the above eq.
 one year ago

farukkBest ResponseYou've already chosen the best response.0
wait let me write what i am getting.
 one year ago

farukkBest ResponseYou've already chosen the best response.0
\[\frac{ \sec^3x }{ 3 } secx +c = answer ???\]
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
it should be, yeah. :) you can always check your answers with wolfram.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
http://www.wolframalpha.com/input/?i=integrate%20tan%5E3%20x%20sec%20x%20dx&t=crmtb01
 one year ago

farukkBest ResponseYou've already chosen the best response.0
but answer is differnet in my book?
 one year ago

farukkBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ 3 }(secxtan^2x2secx)+c =answer ( book)\]
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
just take sec x out. and also, the integral, \(\int du=u\)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
\(\int (u^21) du = \int u^2 du  \int du\) following the integral, it becomes, \(\frac{u^3}{3}  u+C\)
 one year ago

farukkBest ResponseYou've already chosen the best response.0
i have already written this.Scroll up a bit
 one year ago

farukkBest ResponseYou've already chosen the best response.0
i just wanna get the same answer as in my book...:)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
oh. unfortunately, either your book's wrong, or the computer integration is wrong. because wolfram agrees with you....lol
 one year ago

farukkBest ResponseYou've already chosen the best response.0
yeah i have seen there but what you think about book's anwer is that wrong,there is really less possibilit though and yes lolz?
 one year ago

farukkBest ResponseYou've already chosen the best response.0
i have three books with same answer(:
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
lol i dun think wolfram's wrong, in anycase....lol
 one year ago

farukkBest ResponseYou've already chosen the best response.0
if we solve this by ''integration by parts'' rather tha by substituion?
 one year ago

farukkBest ResponseYou've already chosen the best response.0
\[\int\limits uvdx= u \int\limits vdx  \int\limits(\int\limits vdx) u'dx\]
 one year ago

ShadowysBest ResponseYou've already chosen the best response.3
same way, but this time i still let du=d sec x tan x dx. and v=sec^2 1 sorry, gtg, but i think you'll still get the same answer...(or that's what my calc tells me)
 one year ago

farukkBest ResponseYou've already chosen the best response.0
@Shadowys okay i think my all books are wrong this time.(there is no benifit or edge having more than one books) :)!
 one year ago

farukkBest ResponseYou've already chosen the best response.0
thanks i took your time
 one year ago
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