anonymous
  • anonymous
find the integral of tan^5(x)sec^3(x) dx how do I do this? thank you!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1442281422437:dw|
Astrophysics
  • Astrophysics
Since the powers are both you can strip one tangent or one secant and use the trig identity \[\tan^2x+1=\sec^2x\]
anonymous
  • anonymous
okay!! so what happens next? :/

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More answers

Astrophysics
  • Astrophysics
You can make a u substitution, have you tried anything?
anonymous
  • anonymous
so u = tan^5(x)sec^3(x) and dv = dx ?
anonymous
  • anonymous
ohhh okay so what would be my next step? :/ sorry, integrals confuse me a lot :(
beginnersmind
  • beginnersmind
You'll need to remember that (secx)' = secxtanx
Astrophysics
  • Astrophysics
You should try messing with it yourself @iheartfood
Astrophysics
  • Astrophysics
oh wait
Astrophysics
  • Astrophysics
lets do u = secx then du = secxtanxdx better to do this \[\int\limits \tan^4xsec^2x secxtanx dx\]
Astrophysics
  • Astrophysics
Save some time this way
anonymous
  • anonymous
Yea astrophysics your second way is better.... then invert the relationship you gave to put it all in terms of sec(x)
Astrophysics
  • Astrophysics
Yup, it just goes to show, there's many ways to approach these :P
thomas5267
  • thomas5267
\[ \int\tan^5(x)\sec^3(x)\,dx\\ \tan^2(x)+1=\sec^2(x)\\ \int\left(\sec^2(x)-1\right)^2(x)\sec^3(x)\tan(x)\,dx\\ u=\sec(x)\\ du=\sec(x)\tan(x)\,dx\\ \int\left(u^2-1\right)u^2\,du \] A bit late lol.
Astrophysics
  • Astrophysics
Hehe
anonymous
  • anonymous
ooh okay haha thank you both!! and so would i sub in now?
anonymous
  • anonymous
\[ \int tan^4(x)sec^2(x) sec(x)tan(x)dx = \int tan^4(x)sec^2(x) d(sec(x)= \int (1-sec^2(x))^2sec^2(x) d(sec(x) \] Then change variables to u=sec(x) \[ \int (1-u^2)^2u^2du\]
anonymous
  • anonymous
@thomas5267 ?
anonymous
  • anonymous
Darn it got cut off... well the only change between the second and third step was subbing in the fact that tan^4=(tan^2)^2 and then tan^2=1-sec^2
Astrophysics
  • Astrophysics
You can integrate it now from \[\int\limits (u^2-1)^2 u^2 du\]
anonymous
  • anonymous
^ :D
thomas5267
  • thomas5267
Either integration by parts or multiply the whole thing out.
anonymous
  • anonymous
so i get integral of (u^4-1)(u^2)du --> integral of u^6-u^2 du so then i get ((u^7)/7) - ((u^3)/3) du ?
Astrophysics
  • Astrophysics
Hard to tell what you're doing \[\int\limits (u^2-1)^2u^2 du \implies \int\limits (u^4-2u^2+1)u^2 du\]
anonymous
  • anonymous
Yes the equations form is by far superior... btw can I get one medal pretty please though astro deserves the rest since he is finishing it off
anonymous
  • anonymous
Equations *button* form :D
Astrophysics
  • Astrophysics
I gave you one, sorry thomas took one from you so everyone would have one haha :P
anonymous
  • anonymous
TY <3
anonymous
  • anonymous
lol
anonymous
  • anonymous
ohhh :O wait so do i distribute next?
Astrophysics
  • Astrophysics
Yes, there still is the u^2
anonymous
  • anonymous
Oh wow I didnt see Thomas had beaten me to the punch he does deserve the medal there... I was posting in another problem
anonymous
  • anonymous
Whoopsie :/
anonymous
  • anonymous
so then i get integral of u^6 - 2u^4 + u^2 du ?
Astrophysics
  • Astrophysics
Yes
anonymous
  • anonymous
okay! and so i integrate and get this? |dw:1442282862540:dw|
Astrophysics
  • Astrophysics
You drop the integral sign and du when you integrate, now you can plug back your u substitution and add a +C
anonymous
  • anonymous
ohh okay i see so i would end up with this? |dw:1442282938959:dw|
Astrophysics
  • Astrophysics
\[\large \checkmark\]
anonymous
  • anonymous
ooh yay! and so I'm done? :O
Astrophysics
  • Astrophysics
Yes
anonymous
  • anonymous
yay!! thank you so much!!:D
Astrophysics
  • Astrophysics
Np @thomas5267 and @PlasmaFuzer did great work to, collaborative, which makes it funner!
thomas5267
  • thomas5267
So Mr Astrophysics has beaten me has beaten PlasmaFuzer? An ordering on OpenStudy lol?
anonymous
  • anonymous
Haha makes sense @thomas5267 sorry for trying to steal your thunder im bouncing around in too many questions I guess
anonymous
  • anonymous
Definitely agree @Astrophysics... Its also more fun to do harder problems :D
thomas5267
  • thomas5267
Sometimes it is really fun to race with others and spit out the correct answer fastest!
Astrophysics
  • Astrophysics
Hehe agreed!
anonymous
  • anonymous
Yea totally agree it adds to the excitement since many questions can be fairly dull
thomas5267
  • thomas5267
Most question here are quite dull. However I am not that good to answer questions on Math StackExchange.
Astrophysics
  • Astrophysics
There can be good questions, it's just hard to find the ones you're interested in, and most other sites I've encountered aren't as kind as the people here, very discouraging at times.

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