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Math Help - integral question..

  1. #1
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    integral question..

    <br />
\int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\<br />
    <br />
z=\frac{1}{2}\tan x\\<br />
    <br />
dz=\frac{1}{2}\frac{dx}{\cos ^2 x}<br />
    what to do next??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    <br />
\int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\<br />
    <br />
z=\frac{1}{2}\tan x\\<br />
    <br />
dz=\frac{1}{2}\frac{dx}{\cos ^2 x}<br />
    what to do next??
    Read this thread: http://www.mathhelpforum.com/math-he...functions.html
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  3. #3
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    i am not using sec
    only sin cos tan cot

    its very complicated and with sec
    so if i will put sec=1/cos then i ll get a very complicated expression
    is there an easier way??
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  4. #4
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    is there an easier way??
    You could try a hyperbolic substitution. In this example we're looking to reduce the sum of two squares to a single square, z = \frac{1}{2}\sinh \theta will work here.

    z = \frac{1}{2}\sinh \theta \implies dz = \frac{1}{2}\cosh\theta\, d\theta

    So

    I = \int\frac{\frac{1}{2}\cosh\theta \, d\theta}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\f  rac{1}{2}\right)^2\sinh\theta}} = \int d\theta = \theta + C

    We need to write this in terms of our original variable so

    z = \frac{1}{2}\sinh\theta \implies \theta = \sinh^{-1}(2z)

    So

    I = \sinh^{-1}(2z) + C

    This is also the answer maxima gives to the integral
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  5. #5
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    i cant use hyperbolic function
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  6. #6
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    Why not?

    Well if that's really the case then the substitution reduces your original integral to

    \int \sec x \, d x

    once you figure that out, draw a right angled triangle with \tan x = 2z and use it to work out what any other trig functions in your answer are in terms of z
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  7. #7
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    Quote Originally Posted by transgalactic View Post
    i am not using sec
    only sin cos tan cot

    its very complicated and with sec
    so if i will put sec=1/cos then i ll get a very complicated expression
    is there an easier way??
    What do you mean you're not using sec?

    What does the stuff in red say?:

    Quote Originally Posted by transgalactic View Post
    [snip]
    \int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}= {\color{red}\int \frac{dx}{\cos x}}
    [snip]
    What to do next?
    Unless my eyes need testing, I see sec. In which case, I have directed you to what to do next.
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    <br />
\int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\<br />
    <br />
z=\frac{1}{2}\tan x\\<br />
    <br />
dz=\frac{1}{2}\frac{dx}{\cos ^2 x}<br />
    what to do next??
    This question may have been posted and answered at other websites that provide maths help.
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  9. #9
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    i didnt get an answer in a thread posted at another website

    so help is needed
    Last edited by mr fantastic; March 3rd 2009 at 03:36 PM.
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  10. #10
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  11. #11
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    Quote Originally Posted by transgalactic View Post
    i didnt get an answer in this thread
    S.O.S. Mathematics CyberBoard :: View topic - integral question..

    so help is needed
    I gave you the answer in post #2. In post #3 you rejected what I posted. Now you're saying that you need the exact help I gave you that you had previously rejected.

    Thread closed.
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