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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

farukk
 one year 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
 one year 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
 one year 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) :)!

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