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Math Help - Help me with my calculus homework

  1. #1
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    Help me with my calculus homework

    can you help me with this calculus problem?

    here it is

    integral of dy/sqrt[(11*y^2)+5]

    i use algebraic substitution but i cant come up with an answer.. thnx in advance..
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  2. #2
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    Krizalid's Avatar
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    Okay, I'll help ya:

    First off, put y=\frac{\sqrt5}{\sqrt{11}}t, and the integrand (with no constants) will be \frac1{\sqrt{t^2+1}}.

    Wonder how to integrate that? I know four methods, but the most common one is the straightforward trig. substitution.
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  3. #3
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    oh.. i did not get that one.. can u explain the process.. how did you get off the constant??

    i tried it with substitution but i just cant get it.. it will just re turn to the same equation but the sqrt is at the numerator..

    ot:
    how can u do the image something up there?? the image of the Y=sqrt of5/sqrt of 11 *t??
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  4. #4
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    The imagination it's quite simple: put y=\frac{\sqrt5}{\sqrt{11}}t (I'm not going to differentiate this, because I want you to try to see what I'm doing), and then \frac{1}{\sqrt{11y^{2}+5}}=\frac{1}{\sqrt{11\left( \frac{\sqrt{5}}{\sqrt{11}}t \right)^{2}+5}}=\frac{1}{\sqrt{5t^{2}+5}}=\frac{1}  {\sqrt{5}}\cdot \frac{1}{\sqrt{t^{2}+1}}. Does this make sense? Does it?
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  5. #5
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    oh.. im sorry but where did you get the sqrt of 5/sqrt of 11*t

    did you do something to it?? or you just let y be equal to that constant??

    i mean i get the lat part cause its simple algebra but the first part??
    Last edited by magalingakosamath; August 8th 2009 at 07:16 PM.
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  6. #6
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    ok this is easy to differetiate
    y=\frac{\sqrt5}{\sqrt{11}}t

    then after this one, the derivative of y

    dy=\frac{\sqrt5}{\sqrt{11}}
    am i right.. dont we need to substitute dy in the numerator since the original equation goes like this

    \frac{dy}{\sqrt{11y^{2}+5}}

    so im really wonderin were did you get
    y=\frac{\sqrt5}{\sqrt{11}}t<br />

    cant we let
    t={\sqrt{11y^2+5}}

    then it will become
    y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=<br />
\frac{t-\sqrt{5}}{\sqrt{11}}

    then getting dy we will just use the formula integral of u/v is vdu-udv all over v^2 ayt?? that's the first method i did but i got confused.. then when you let y be equal to sqrt of 5 over 11 i also got confused..
    Last edited by magalingakosamath; August 8th 2009 at 07:40 PM.
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  7. #7
    Member eXist's Avatar
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    Quote Originally Posted by Krizalid View Post
    y=\frac{\sqrt5}{\sqrt{11}}t,
    What he's doing here is straight forward. He's letting y = that expression, and then substituting it into the equation to make it easier to integrate. Looking at the new integral after substitution, can you integrate that?
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  8. #8
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    yes i can, by trigonometric substitution..

    ive already solved it but using algebraic substitution.. its simple.. but i dont know if its correct....

    here is what i did.. from the original equation

    i let
    t={\sqrt{11y^2+5}}

    then
    y=\frac{\sqrt{t^2-5}}{\sqrt{11}}=<br />
\frac{t-\sqrt{5}}{\sqrt{11}}

    i get the dy.. and it become
    dy=\frac{\sqrt{11}}{11}dt

    then i substitute the values of dy and t to the original equation and i get..
    \frac{\sqrt{11}}{11}\int{\frac{dt}{t}}

    then the answer will be

    \frac{\sqrt{11}}{11}\ln{t}+c

    am i correct??
    Last edited by magalingakosamath; August 8th 2009 at 10:48 PM.
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  9. #9
    Member eXist's Avatar
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    Well take the derivative of your answer, do you get what you started with?
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  10. #10
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    yes i get it.. i let the denomenator to be equal to "t" ,then i get the value of "y" in terms of "t".. then get the derivative of "y"...

    so the original equation is
    \int{\frac{dy}{\sqrt{11y^{2}+5}}}

    i just substitute the value of dy and the denomenator so that i can integrate it easily.. then i came up with my answer.. so y do i still need to differentiate my answer??
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  11. #11
    Member eXist's Avatar
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    I meant you can differentiate it to check to see if your answer is correct right? That's all I meant. You DO NOT have to differentiate your answer.
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  12. #12
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    oh im so sorry bout that..

    i get it., hehehe.. its the same.. but it has an excess of sqrt of 11.. maybe thats the constant c.. am i right??
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