A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

find the integral of tan^5(x)sec^3(x) dx how do I do this? thank you!

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1442281422437:dw|

  2. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Since the powers are both you can strip one tangent or one secant and use the trig identity \[\tan^2x+1=\sec^2x\]

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay!! so what happens next? :/

  4. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    You can make a u substitution, have you tried anything?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so u = tan^5(x)sec^3(x) and dv = dx ?

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhh okay so what would be my next step? :/ sorry, integrals confuse me a lot :(

  7. beginnersmind
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You'll need to remember that (secx)' = secxtanx

  8. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    You should try messing with it yourself @iheartfood

  9. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    oh wait

  10. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    lets do u = secx then du = secxtanxdx better to do this \[\int\limits \tan^4xsec^2x secxtanx dx\]

  11. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Save some time this way

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yea astrophysics your second way is better.... then invert the relationship you gave to put it all in terms of sec(x)

  13. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Yup, it just goes to show, there's many ways to approach these :P

  14. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[ \int\tan^5(x)\sec^3(x)\,dx\\ \tan^2(x)+1=\sec^2(x)\\ \int\left(\sec^2(x)-1\right)^2(x)\sec^3(x)\tan(x)\,dx\\ u=\sec(x)\\ du=\sec(x)\tan(x)\,dx\\ \int\left(u^2-1\right)u^2\,du \] A bit late lol.

  15. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Hehe

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ooh okay haha thank you both!! and so would i sub in now?

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \int tan^4(x)sec^2(x) sec(x)tan(x)dx = \int tan^4(x)sec^2(x) d(sec(x)= \int (1-sec^2(x))^2sec^2(x) d(sec(x) \] Then change variables to u=sec(x) \[ \int (1-u^2)^2u^2du\]

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @thomas5267 ?

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  20. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    You can integrate it now from \[\int\limits (u^2-1)^2 u^2 du\]

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^ :D

  22. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Either integration by parts or multiply the whole thing out.

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 ?

  24. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Hard to tell what you're doing \[\int\limits (u^2-1)^2u^2 du \implies \int\limits (u^4-2u^2+1)u^2 du\]

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Equations *button* form :D

  27. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    I gave you one, sorry thomas took one from you so everyone would have one haha :P

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    TY <3

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhh :O wait so do i distribute next?

  31. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes, there still is the u^2

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh wow I didnt see Thomas had beaten me to the punch he does deserve the medal there... I was posting in another problem

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Whoopsie :/

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so then i get integral of u^6 - 2u^4 + u^2 du ?

  35. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay! and so i integrate and get this? |dw:1442282862540:dw|

  37. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    You drop the integral sign and du when you integrate, now you can plug back your u substitution and add a +C

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohh okay i see so i would end up with this? |dw:1442282938959:dw|

  39. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \[\large \checkmark\]

  40. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ooh yay! and so I'm done? :O

  41. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yay!! thank you so much!!:D

  43. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Np @thomas5267 and @PlasmaFuzer did great work to, collaborative, which makes it funner!

  44. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So Mr Astrophysics has beaten me has beaten PlasmaFuzer? An ordering on OpenStudy lol?

  45. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha makes sense @thomas5267 sorry for trying to steal your thunder im bouncing around in too many questions I guess

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Definitely agree @Astrophysics... Its also more fun to do harder problems :D

  47. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sometimes it is really fun to race with others and spit out the correct answer fastest!

  48. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Hehe agreed!

  49. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yea totally agree it adds to the excitement since many questions can be fairly dull

  50. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Most question here are quite dull. However I am not that good to answer questions on Math StackExchange.

  51. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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.

  52. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.