anonymous
  • anonymous
integrate tan^3 xsec x dx...!
Mathematics
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anonymous
  • anonymous
integrate tan^3 xsec x dx...!
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
\[\int\limits \tan^3 x secx dx\]
anonymous
  • anonymous
what to do?
experimentX
  • experimentX
\[ \tan^3 x \sec x = \tan x \sec x ( \sec^2 x - 1)\] seems like it will get the same

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anonymous
  • anonymous
yeah ..next step
anonymous
  • anonymous
note that d(sec x) =sec x tan x dx
anonymous
  • anonymous
yes all right
anonymous
  • anonymous
what's next..
anonymous
  • anonymous
let u = sec x then proceed. du= sec x tan x dx so it all becomes \(\int (u^2 - 1) du\)
anonymous
  • anonymous
dear are you solving integral by parts?
anonymous
  • anonymous
or by substituion?
anonymous
  • anonymous
it will become like this i guess u^3/3 - u +c
anonymous
  • anonymous
\( \int (\sec^2-1) \tan x \sec x dx\) =\( \int (\sec^2-1) d(\sec x)\) =\( \int (u^2-1) du\), where u=\sec x\) by substitution.
anonymous
  • anonymous
yes next step..
anonymous
  • anonymous
eh? \(\int u^n du= \frac{u^{n+1}}{n+1} +C\)
anonymous
  • anonymous
after applying we ll get as i above wrote ?
anonymous
  • anonymous
you'll get the answer....from integrating the above eq.
anonymous
  • anonymous
wait let me write what i am getting.
anonymous
  • anonymous
\[\frac{ \sec^3x }{ 3 }- secx +c = answer ???\]
anonymous
  • anonymous
it should be, yeah. :) you can always check your answers with wolfram.
anonymous
  • anonymous
wolfram.?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integrate%20tan%5E3%20x%20sec%20x%20dx&t=crmtb01
anonymous
  • anonymous
but answer is differnet in my book?
anonymous
  • anonymous
\[\frac{ 1 }{ 3 }(secxtan^2x-2secx)+c =answer ( book)\]
anonymous
  • anonymous
anonymous
  • anonymous
just take sec x out. and also, the integral, \(\int du=u\)
anonymous
  • anonymous
did't get..
anonymous
  • anonymous
\(\int (u^2-1) du = \int u^2 du - \int du\) following the integral, it becomes, \(\frac{u^3}{3} - u+C\)
anonymous
  • anonymous
i have already written this.Scroll up a bit
anonymous
  • anonymous
i just wanna get the same answer as in my book...:)
anonymous
  • anonymous
oh. unfortunately, either your book's wrong, or the computer integration is wrong. because wolfram agrees with you....lol
anonymous
  • anonymous
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?
anonymous
  • anonymous
i have three books with same answer(:
anonymous
  • anonymous
lol i dun think wolfram's wrong, in anycase....lol
anonymous
  • anonymous
if we solve this by ''integration by parts'' rather tha by substituion?
anonymous
  • anonymous
\[\int\limits uvdx= u \int\limits vdx - \int\limits(\int\limits vdx) u'dx\]
anonymous
  • anonymous
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)
anonymous
  • anonymous
@Shadowys okay i think my all books are wrong this time.(there is no benifit or edge having more than one books) :)!
anonymous
  • anonymous
thanks i took your time

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