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CatLove9

  • 3 years ago

Integrate. sqrt(1+x^2)/x

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  1. CatLove9
    • 3 years ago
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    I think it should be \[\int\limits_{}^{} 1/\sin(x)\cos^2x\]

  2. Mimi_x3
    • 3 years ago
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    Or start from \[\int\frac{\sqrt{1+x^{2}}}{x} dx\]

  3. CatLove9
    • 3 years ago
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    You could started from there, I got up to 1/sinxcos^2x

  4. Mimi_x3
    • 3 years ago
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    Assuming that you did the other part correct then.. \[\int \frac{1}{sinxcos^2{x}}\] wait..are you sure about that though?

  5. lgbasallote
    • 3 years ago
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    |dw:1342236036378:dw| \[\sec \theta = \sqrt{1+ x^2}\] \[\tan \theta = x\] \[\sec^2 \theta d\theta = dx\] \[\int \frac{\sqrt{1+x^2}}{x}dx \implies \frac{\sec \theta(\sec^2 \theta d\theta )}{\tan \theta}\]

  6. CatLove9
    • 3 years ago
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    I'm sure that \[\int\limits_{}^{} \sec^3x/\tan(x)\]

  7. CatLove9
    • 3 years ago
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    That turns into 1/cos^2xsinx

  8. lgbasallote
    • 3 years ago
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    it does?

  9. lgbasallote
    • 3 years ago
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    i mean yeah it does

  10. CatLove9
    • 3 years ago
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    I know I can't do u sub.

  11. lgbasallote
    • 3 years ago
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    there's only one thing you can do...integration by parts

  12. lgbasallote
    • 3 years ago
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    \[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \csc x \sec^2 x dx\]

  13. CatLove9
    • 3 years ago
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    But can I integrate that?

  14. lgbasallote
    • 3 years ago
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    well i was just spitting out ideas....

  15. CatLove9
    • 3 years ago
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    That's why I didn't change it to that because I knew there was nothing to do there.

  16. lgbasallote
    • 3 years ago
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    well i know i can make 1 = sin^2 x + cos^2 x

  17. lgbasallote
    • 3 years ago
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    \[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \frac{ \sin^2 x + \cos^2 x}{\sin x \cos^2 x}dx\]

  18. CatLove9
    • 3 years ago
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    OH! That's good

  19. lgbasallote
    • 3 years ago
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    \[\int \frac{\sin^2 x}{\sin x \cos ^2 x}dx + \int \frac{\cos^2 x}{\sin x \cos^2 x}dx\]

  20. lgbasallote
    • 3 years ago
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    yup saw the answer now :D i'll let you do the good stuff ;)

  21. CatLove9
    • 3 years ago
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    Ah, yes, yes, I see it too.

  22. lgbasallote
    • 3 years ago
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    nice! congrats

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