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MathSofiya

\[\int cosxsec^2xdx\] 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?

  • one year ago
  • one year ago

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  1. Valpey
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    Asking cos*sec^2, not cos/sec^2

    • one year ago
  2. bahrom7893
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    Oh, sorry: Cosx/Cos^2x = 1/Cosx = Secx

    • one year ago
  3. bahrom7893
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    Well actually you'd keep it as 1/Cox, and use integration by parts

    • one year ago
  4. MathSofiya
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    wait a second \[\int cosx\frac 1{cos^2x}dx?\]

    • one year ago
  5. Valpey
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    Yes.

    • one year ago
  6. bahrom7893
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    Sorry for the confusion.. Let's start over: Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx

    • one year ago
  7. MathSofiya
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    \[\int\frac 1{cosx}dx\]

    • one year ago
  8. Valpey
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    \[\int{\cos{x}*\frac{1}{1-\sin^2{x}}}\ dx\]

    • one year ago
  9. bahrom7893
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    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

    • one year ago
  10. MathSofiya
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    why did you change it? wouldn't it make more sense to cancel one cosine out?

    • one year ago
  11. bahrom7893
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    No

    • one year ago
  12. Valpey
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    Let u = sin(x) du = cos(x)dx \[\int{\frac{du}{1-u^2}}\]

    • one year ago
  13. bahrom7893
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    Actually this would be the easiest option, imo: Cos^2(u) = (1/2)+(Cos(2u))/2

    • one year ago
  14. bahrom7893
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    ok let Valpey, finish this, otherwise you'll get confused.

    • one year ago
  15. MathSofiya
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    sure

    • one year ago
  16. MathSofiya
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    do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]

    • one year ago
  17. MathSofiya
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    \[\frac {1+u^2}{1+u^2}\]

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

    • one year ago
  19. MathSofiya
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    makes sense, thank you!

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

    • one year ago
  21. Valpey
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    *\[\frac{1}{2}\Big(\log(1+\sin x)-\log(1-\sin x)\Big)\]

    • one year ago
  22. Valpey
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    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.

    • one year ago
  23. TuringTest
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    @MathSofiya still want help with\[\int\cos x\sec^2xdx\]?

    • one year ago
  24. TuringTest
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    it seems that everyone here has forgotten the simple trick to this integral...

    • one year ago
  25. MathSofiya
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    see there had to be a simpler way

    • one year ago
  26. TuringTest
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    \[\cos x\sec^2x=\cos x\cdot\frac1{\cos^2 x}=\frac1{\cos x}=\sec x\]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)

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

    • one year ago
  28. TuringTest
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    that is the standard approach for this well-known integral

    • one year ago
  29. MathSofiya
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    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.

    • one year ago
  30. MathSofiya
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    but yes it makes sense now I see it.

    • one year ago
  31. TuringTest
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    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 :)

    • one year ago
  32. MathSofiya
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    I shall memorize this integral =P (honestly I will) And as always, Thanks Turing!

    • one year ago
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