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Math Help - Got this one in my test today...

  1. #1
    Bar0n janvdl's Avatar
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    Got this one in my test today...

    Hello everyone

    I got this one in my test today and would like to know what the answer should have been, please. (EDIT: I used partial integration on it)

    Thanks in advance.

    \int \frac{1 - t}{\sqrt{9 - t^2}}
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  2. #2
    Senior Member polymerase's Avatar
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    make the substitution u^2=9-t^2

    You then end up with this, -\int \frac{1-\sqrt{9-u^2}}{\sqrt{9-u^2}}=u-\arcsin \left(\frac{u}{3}\right)+C

    Thus \int \frac{1-t}{\sqrt{9 - t^2}}=\sqrt{9-t^2}-\arcsin\left(\frac{\sqrt{9-t^2}}{3}\right)+C
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  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
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    Hello janvdl,

    Substitute t = 3\sin x

    We get \displaystyle \int (1 - 3\sin x) \, dx = x + 3\cos x + C = \sin^{-1} \frac{t}{3} + 3\sqrt{1 - t^2}+C

    Whats your method?
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by polymerase View Post
    make the substitution u^2=9-t^2

    You then end up with this, -\int \frac{1-\sqrt{9-u^2}}{\sqrt{9-u^2}}=u-\arcsin \left(\frac{u}{3}\right)+C

    Thus \int \frac{1-t}{\sqrt{9 - t^2}}=\sqrt{9-t^2}-\arcsin\left(\frac{\sqrt{9-t^2}}{3}\right)+C
    Hehe okay, i got it wrong then... Thanks!
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Hello janvdl,

    Substitute t = 3\sin x

    We get \displaystyle \int (1 - 3\sin x) \, dx = x + 3\cos x + C = \sin^{-1} \frac{t}{3} + 3\sqrt{1 - t^2}+C

    Whats your method?
    Thanks for your reply, but I don't understand why you substituted using Sin(x). Maybe that is a more advanced method?
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  6. #6
    Senior Member polymerase's Avatar
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    Quote Originally Posted by janvdl View Post
    Thanks for your reply, but I don't understand why you substituted using Sin(x). Maybe that is a more advanced method?
    That's trig. substitution. By letting t=3\sin x, you end up with \sqrt{9-9\sin^2 x}=3\sqrt{1-\sin^2 x}=3\sqrt{\cos^2 x}=3\cos x Get it?
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by polymerase View Post
    That's trig. substitution. By letting t=3\sin x, you end up with \sqrt{9-9\sin^2 x}=3\sqrt{1-\sin^2 x}=3\sqrt{\cos^2 x}=3\cos x Get it?
    Got it Thanks.
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by janvdl View Post
    Thanks for your reply, but I don't understand why you substituted using Sin(x). Maybe that is a more advanced method?
    Not really,if you did learn substitution, I think you should be familiar with this.

    Whenever you see \sqrt{a^2 - x^2} there is an itching to get rid of the square root. And the standard substitution is x = a\sin u. \sqrt{a^2 - x^2} neatly simplifies to a\cos u. Moreover dx = a\cos u \, du. Everything neatly becomes trigonometric
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  9. #9
    Bar0n janvdl's Avatar
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    Right, I think I got it now Thanks again guys!
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