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anonymous

  • 5 years ago

Help with integrals

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  1. anonymous
    • 5 years ago
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    what do you need help with?

  2. anonymous
    • 5 years ago
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    \[\int\limits_{}^{}\frac{dx}{\sin^2x}\]

  3. anonymous
    • 5 years ago
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    I got stuck here :\[2\int\limits_{}^{}\frac{dx}{1-\cos(2x)}\]

  4. anonymous
    • 5 years ago
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    hold on let me write it out ok?

  5. anonymous
    • 5 years ago
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    but btw the 1-cos(2x) use trig identitites.

  6. anonymous
    • 5 years ago
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    sure. thanks

  7. anonymous
    • 5 years ago
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    So how is it?

  8. anonymous
    • 5 years ago
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    \[\int\limits 1\div \sin x ^{2} dx = x / sinx ^{2}\]

  9. anonymous
    • 5 years ago
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    Nah.

  10. anonymous
    • 5 years ago
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    -1/tan x sorry wrong answer.

  11. anonymous
    • 5 years ago
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    How did you get it?

  12. anonymous
    • 5 years ago
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    It's normally assumed from the fact \[\frac{\mathbb{d}}{\mathbb{d}x} \cot x = -cosec^2x\] .... but if you REALLY want to prove it, try the sub \[t = \tan \frac{x}{2} \] Haven't tried it, but it almost always works. May be some easy way I'm missing.

  13. anonymous
    • 5 years ago
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    Note, proving the differentiation version, if you deem that sufficient, is far easier. But if you did not know the result, the sub I guess would be OK.

  14. anonymous
    • 5 years ago
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    moolean, i'd stick with what hes saying, i think he understand integrals better then i do. im nto perfect at it yet, i was just trying to help, but he seems to rlly know what he's talking about.

  15. anonymous
    • 5 years ago
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    \[\text{Let } t = \frac{\tan{x}}{2} \implies \frac{\mathbb{d}x}{\mathbb{d}t} = \frac{2}{1+t^2} \] It quickly follows that: \[\sin x = \frac{2t}{1+t^2}\] \[\int \frac{1}{sin^2x} \mathbb{d}x = \int \left(\frac{1+t^2}{2t}\right)^2\cdot \frac{2}{1+t^2} \mathbb{d}t = \int \left(\frac{1+t^2}{2t^2}\right)\cdot \mathbb{d}t \] \[\int \left(\frac{1+t^2}{2t^2}\right)^2\cdot \mathbb{d}t = \frac{t^2 - 1}{2t} \] Which, from double angle formulae, is equal to the required result.

  16. anonymous
    • 5 years ago
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    It did work, after all :D

  17. anonymous
    • 5 years ago
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    gj Newton :P that's impressive.

  18. anonymous
    • 5 years ago
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    Thanks - I needed to practice it anyway. NOTE there should be no squared on the fraction on the last line. some dodgy copy/pasting from above there :(

  19. anonymous
    • 5 years ago
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    THAT;S A GREAT HELP. THANKS!

  20. anonymous
    • 5 years ago
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    Ugh, the sub is \[ t = \tan \frac{x}{2} \] So many typos :/ But you're welcome.

  21. anonymous
    • 5 years ago
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    geez i should say thank you too, u taught me something and i was only trying to help :P

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