Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

MathSofiya Group Title

\[\int cosxsec^2xdx\] I forgot how to do this integral manually this is what wolf had to say http://www.wolframalpha.com/input/?i=integral+of+cosxsec^2x how how do I do it manually?

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Asking cos*sec^2, not cos/sec^2

    • 2 years ago
  2. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, sorry: Cosx/Cos^2x = 1/Cosx = Secx

    • 2 years ago
  3. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Well actually you'd keep it as 1/Cox, and use integration by parts

    • 2 years ago
  4. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    wait a second \[\int cosx\frac 1{cos^2x}dx?\]

    • 2 years ago
  5. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes.

    • 2 years ago
  6. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry for the confusion.. Let's start over: Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx

    • 2 years ago
  7. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int\frac 1{cosx}dx\]

    • 2 years ago
  8. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int{\cos{x}*\frac{1}{1-\sin^2{x}}}\ dx\]

    • 2 years ago
  9. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Well that I just had memorized, but actually i bet you don't simplify it down to 1/Cos, I was probably wrong again, you keep it as Cos/Cos^2 and then do what Valpey told you to

    • 2 years ago
  10. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    why did you change it? wouldn't it make more sense to cancel one cosine out?

    • 2 years ago
  11. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    No

    • 2 years ago
  12. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Let u = sin(x) du = cos(x)dx \[\int{\frac{du}{1-u^2}}\]

    • 2 years ago
  13. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually this would be the easiest option, imo: Cos^2(u) = (1/2)+(Cos(2u))/2

    • 2 years ago
  14. bahrom7893 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok let Valpey, finish this, otherwise you'll get confused.

    • 2 years ago
  15. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    sure

    • 2 years ago
  16. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]

    • 2 years ago
  17. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac {1+u^2}{1+u^2}\]

    • 2 years ago
  18. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \int \sec x dx = \log \left | \sec x + \tan x\right | = \log \left| \tan \left( {x \over 2} + {\pi \over 4}\right)\right | \] wikipedia says it's quite a problem http://en.wikipedia.org/wiki/Integral_of_the_secant_function

    • 2 years ago
  19. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    makes sense, thank you!

    • 2 years ago
  20. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, duh! My bad.\[\int\frac{1}{1-u^2}du=\int\frac{1}{(1-u)(1+u)}du=\frac{1}{2}\int\frac{1+1}{(1-u)(1+u)}du\]\[=\frac{1}{2}\int\Big(\frac{1+u}{(1-u)(1+u)}+\frac{1-u}{(1-u)(1+u)}\Big)du\]\[=\frac{1}{2}\int\frac{1+u}{(1-u)(1+u)}du+\frac{1}{2}\int\frac{1-u}{(1-u)(1+u)}du=\frac{1}{2}\int\frac{1}{1-u}du+\frac{1}{2}\int\frac{1}{1+u}du\] \[=\frac{1}{2}(-\log(1-u))+\frac{1}{2}(\log(1+u))\] \[\frac{1}{2}\Big(\log(1+\sin x)-\log(1+\sin x)\Big)\]

    • 2 years ago
  21. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    *\[\frac{1}{2}\Big(\log(1+\sin x)-\log(1-\sin x)\Big)\]

    • 2 years ago
  22. Valpey Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh yeah, I need absolute values. \[\frac{1}{2}\Big(\log|1+\sin x|−\log|1−\sin x|\Big)\]Which is also\[\frac{1}{2}\log\frac{|1+\sin x|}{|1−\sin x|}\], but now I'm just cheating off Wikipedia.

    • 2 years ago
  23. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @MathSofiya still want help with\[\int\cos x\sec^2xdx\]?

    • 2 years ago
  24. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    it seems that everyone here has forgotten the simple trick to this integral...

    • 2 years ago
  25. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    see there had to be a simpler way

    • 2 years ago
  26. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\cos x\sec^2x=\cos x\cdot\frac1{\cos^2 x}=\frac1{\cos x}=\sec x\]so\[\int\cos x\sec^2 xdx=\int\sec xdx\]now the great trick for this integral: multiply by\[{\sec x+\tan x\over\sec x+\tan x}\]and watch what happens (yes I know the idea is not obvious, but once you see it try to remember it)

    • 2 years ago
  27. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int\sec x\left({\sec x+\tan x\over\sec x+\tan x}\right)dx=\int{\sec^2x+\sec x\tan x\over\sec x+\tan x}dx\]magically the top is now the derivative of the bottom:\[u=\sec x+\tan x\implies du=\sec^2x+\sec x\tan xdx\]\[\int\frac1udu=\ln|u|+C=\ln|\sec x+\tan x|+C\]

    • 2 years ago
  28. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    that is the standard approach for this well-known integral

    • 2 years ago
  29. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh I see. So is it just a matter of doing many integrals and then coming to this conclusion? I wouldn't have thought of that, and yet again I haven't reviewed my integrals much.

    • 2 years ago
  30. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    but yes it makes sense now I see it.

    • 2 years ago
  31. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I don't think I would have thought of this had I not known this integral either, I'm sure whoever discovered this trick spent quite some time on it. Fortunately, most integral tricks are far more obvious, this is just one of the more esoteric, yet directly effective tricks. There are few integrals that require such a non-obvious algebraic manipulation, most are easier too see. That is why I say, memorize this one, it is a classic and will serve you in the future :)

    • 2 years ago
  32. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I shall memorize this integral =P (honestly I will) And as always, Thanks Turing!

    • 2 years ago
    • Attachments:

See more questions >>>

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.