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farukk
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \tan^3 x secx dx\]

experimentX
 2 years ago
Best 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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.3note that d(sec x) =sec x tan x dx

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

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0dear are you solving integral by parts?

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0it will become like this i guess u^3/3  u +c

Shadowys
 2 years ago
Best 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.

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

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0after applying we ll get as i above wrote ?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.3you'll get the answer....from integrating the above eq.

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0wait let me write what i am getting.

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sec^3x }{ 3 } secx +c = answer ???\]

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.3it should be, yeah. :) you can always check your answers with wolfram.

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

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0but answer is differnet in my book?

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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.3just take sec x out. and also, the integral, \(\int du=u\)

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

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0i have already written this.Scroll up a bit

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0i just wanna get the same answer as in my book...:)

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

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

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0i have three books with same answer(:

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.3lol i dun think wolfram's wrong, in anycase....lol

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0if we solve this by ''integration by parts'' rather tha by substituion?

farukk
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits uvdx= u \int\limits vdx  \int\limits(\int\limits vdx) u'dx\]

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.3same 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)

farukk
 2 years ago
Best 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) :)!
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