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MathSofiya

  • 2 years ago

\[\int cosxsec^2xdx\] 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?

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

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

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

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

  5. Valpey
    • 2 years ago
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    Yes.

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

  7. MathSofiya
    • 2 years ago
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    \[\int\frac 1{cosx}dx\]

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

  9. bahrom7893
    • 2 years ago
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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

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

  11. bahrom7893
    • 2 years ago
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    No

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

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

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

  15. MathSofiya
    • 2 years ago
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    sure

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

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

  18. experimentX
    • 2 years ago
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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

  19. MathSofiya
    • 2 years ago
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    makes sense, thank you!

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

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

  22. Valpey
    • 2 years ago
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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.

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

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

  25. MathSofiya
    • 2 years ago
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    see there had to be a simpler way

  26. TuringTest
    • 2 years ago
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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)

  27. TuringTest
    • 2 years ago
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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:\[u=\sec x+\tan x\implies du=\sec^2x+\sec x\tan xdx\]\[\int\frac1udu=\ln|u|+C=\ln|\sec x+\tan x|+C\]

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

  29. MathSofiya
    • 2 years ago
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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.

  30. MathSofiya
    • 2 years ago
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    but yes it makes sense now I see it.

  31. TuringTest
    • 2 years ago
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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 :)

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

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