Question

integrate tan^3 xsec x dx...!

Answer 1

\[\int\limits \tan^3 x secx dx\]

Answer 2

what to do?

Answer 3

\[ \tan^3 x \sec x = \tan x \sec x ( \sec^2 x - 1)\]
seems like it will get the same

Answer 4

yeah ..next step

Answer 5

note that d(sec x) =sec x tan x dx

Answer 6

yes all right

Answer 7

what's next..

Answer 8

let u = sec x then proceed.
du= sec x tan x dx
so it all becomes \(\int (u^2 - 1) du\)

Answer 9

dear are you solving integral by parts?

Answer 10

or by substituion?

Answer 11

it will become like this i guess u^3/3 - u +c

Answer 12

\( \int (\sec^2-1) \tan x \sec x dx\)
=\( \int (\sec^2-1) d(\sec x)\)
=\( \int (u^2-1) du\), where u=\sec x\)
by substitution.

Answer 13

yes next step..

Answer 14

eh? \(\int u^n du= \frac{u^{n+1}}{n+1} +C\)

Answer 15

after applying we ll get as i above wrote ?

Answer 16

you'll get the answer....from integrating the above eq.

Answer 17

wait let me write what i am getting.

Answer 18

\[\frac{ \sec^3x }{ 3 }- secx +c = answer ???\]

Answer 19

it should be, yeah. :) you can always check your answers with wolfram.

Answer 20

wolfram.?

Answer 21

http://www.wolframalpha.com/input/?i=integrate%20tan%5E3%20x%20sec%20x%20dx&t=crmtb01

Answer 22

but answer is differnet in my book?

Answer 23

\[\frac{ 1 }{ 3 }(secxtan^2x-2secx)+c =answer ( book)\]

Answer 24

@Shadowys

Answer 25

just take sec x out. and also, the integral, \(\int du=u\)

Answer 26

did't get..

Answer 27

\(\int (u^2-1) du = \int u^2 du - \int du\)
following the integral, it becomes, \(\frac{u^3}{3} - u+C\)

Answer 28

i have already written this.Scroll up a bit

Answer 29

i just wanna get the same answer as in my book...:)

Answer 30

oh. unfortunately, either your book's wrong, or the computer integration is wrong. because wolfram agrees with you....lol

Answer 31

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?

Answer 32

i have three books with same answer(:

Answer 33

lol i dun think wolfram's wrong, in anycase....lol

Answer 34

if we solve this by ''integration by parts'' rather tha by substituion?

Answer 35

\[\int\limits uvdx= u \int\limits vdx - \int\limits(\int\limits vdx) u'dx\]

Answer 36

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)

Answer 37

@Shadowys okay i think my all books are wrong this time.(there is no benifit or edge having more than one books) :)!

Answer 38

thanks i took your time