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Math Help - trigonometric substitution

  1. #1
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    trigonometric substitution

    \int \frac{\sqrt {4x^2-16}}{x} dx<br />

    = \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx

    2x = 4 sec t
    x = 2 sec t

    \int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx
    \int \frac {\sqrt {16(sec^2t-1}}{2sect} dx
    \int \frac {4 \sqrt {sec^2t-1}}{2sect} dx

    How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    \int \frac{\sqrt {4x^2-16}}{x} dx<br />

    = \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx

    2x = 4 sec t
    x = 2 sec t

    \int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx
    \int \frac {\sqrt {16(sec^2t-1}}{2sect} dx
    \int \frac {4 \sqrt {sec^2t-1}}{2sect} dx

    How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.
    sec^2t-1=tan^2t

    dx=2sect tant dt
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  3. #3
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    Quote Originally Posted by ione View Post
    sec^2t-1=tan^t
    Thanks. So I finally end up with;

     <br />
2 \int \frac {tan t}{sec t}<br />

    I'm not certain here, but does it become;

     <br />
2 \frac {-ln |cos t|}{ln |sec t + tan t|} + C ?<br />
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  4. #4
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    Hello Archduke01
    Quote Originally Posted by Archduke01 View Post
    \int \frac{\sqrt {4x^2-16}}{x} dx<br />

    = \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx

    2x = 4 sec t
    x = 2 sec t

    \int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx
    \int \frac {\sqrt {16(sec^2t-1}}{2sect} dx
    \int \frac {4 \sqrt {sec^2t-1}}{2sect} dx

    How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.
    You don't need the trig substitution, if you put
    4x^2 - 16 = u^2

     \Rightarrow x\,dx = \tfrac14u\,du

    Then:
    \int \frac{\sqrt {4x^2-16}}{x} dx
    =\int \frac{\sqrt {4x^2-16}}{x^2}\,x\,dx

    =\int\frac{u}{\tfrac14(u^2+16)}\;\tfrac14u\,du


    =\int\frac{u^2}{u^2+16}\;du


    =\int\left(1-\frac{16}{u^2+16}\right)\;du

    Can you complete it now?

    Grandad
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  5. #5
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    I know trig sub isn't necessary, but I've been asked to use it since it's the new topic. Thank you for your time, though.
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    Quote Originally Posted by Archduke01 View Post
    Thanks. So I finally end up with;

     <br />
2 \int \frac {tan t}{sec t}<br />

    I'm not certain here, but does it become;

     <br />
2 \frac {-ln |cos t|}{ln |sec t + tan t|} + C ?<br />
     <br />
2 \int \frac {tan t}{sec t}dx<br />

    dx=2sect\,tant\,dt
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  7. #7
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    Quote Originally Posted by ione View Post
     <br />
2 \int \frac {tan t}{sec t}dx<br />

    dx=2sect\,tant\,dt
    Thanks broseidon. I end up with;

     <br />
2 \int \frac {tan t}{sec t} 2 sect tan t dt<br />
    = 4 \int tan t tan t dt
     <br />
= 4 \int tan^2t dt<br />

    I'm sorry for the continuous questions, but how can I integrate that value? I can't find a rule to integrate tan^2 t
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  8. #8
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    Quote Originally Posted by Archduke01 View Post
    Thanks broseidon. I end up with;

     <br />
2 \int \frac {tan t}{sec t} 2 sect tan t dt<br />
    = 4 \int tan t tan t dt
     <br />
= 4 \int tan^2t dt<br />

    I'm sorry for the continuous questions, but how can I integrate that value? I can't find a rule to integrate tan^2 t
    Use tan^2t=sec^2t-1
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  9. #9
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    So it becomes;

    4 \int (sec^2t + 1) dt
    = 4(tan t + t) + C ?
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  10. #10
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    Quote Originally Posted by Archduke01 View Post
    So it becomes;

    4 \int (sec^2t + 1) dt
    = 4(tan t + t) + C ?
    Write your answer in terms of x
    Last edited by ione; March 25th 2010 at 10:41 AM.
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  11. #11
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    Right so I have;

    4 (tan x + x) + C

    Now I've got to construct that triangle. Since originally my x = 2 sec t, it becomes sec t = x/2. If we're looking for the cos it becomes cos t = 2/x. So on the triangle, the hypotenuse is x while the two sides are 2 and \sqrt {x^2-4}.

    But it's wrong apparently...
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  12. #12
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    Quote Originally Posted by Archduke01 View Post
    Right so I have;

    4 (tan x + x) + C

    Now I've got to construct that triangle. Since originally my x = 2 sec t, it becomes sec t = x/2. If we're looking for the cos it becomes cos t = 2/x. So on the triangle, the hypotenuse is x while the two sides are 2 and \sqrt {x^2-4}.

    But it's wrong apparently...
    I find it helpful to draw a right triangle when I begin using trig substitution

    Label one of the acute angles t

    Since , label the hypotenuse 2x and the adjacent side 4

    Then the opposite side will be



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  13. #13
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    Quote Originally Posted by ione View Post
    I find it helpful to draw a right triangle when I begin using trig substitution

    Label one of the acute angles t

    Since , label the hypotenuse 2x and the adjacent side 4

    Then the opposite side will be



    I'm supposed to go to my integrated equation which is 4 (tan x + x) + C.

    Using the triangle, I can now replace the tan x with numbers. So my final answer is;

     <br />
4( \frac {\sqrt{4x^2-16}}{4} + x) + C<br />

    Unfortunately, it's apparently wrong and even after double checking I can't figure out where I messed up.
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  14. #14
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    Quote Originally Posted by Archduke01 View Post
    So it becomes;

    4 \int (sec^2t + 1) dt
    = 4(tan t + t) + C ?
    4 \int (sec^2t - 1) dt

    = 4(tan t - t) + C

    =4tan t - 4t + C

    =\sqrt{4x^2-16} - 4 arcsec(\frac{x}{2}) + C
    Last edited by ione; March 25th 2010 at 01:47 PM.
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  15. #15
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    Hello Archduke01
    Quote Originally Posted by Archduke01 View Post
    So it becomes;

    4 \int (sec^2t + 1) dt
    = 4(tan t + t) + C ?
    You have a sign wrong.

    4\int \tan^2t\;dt
    =4\int(\sec^2t -1)\;dt

    =4(\tan t - t) + c


    = 4(\sqrt{\sec^2t-1}-t) + c


    = 4\big(\sqrt{\tfrac14x^2-1}-\text{arcsec}(\tfrac12x)\big)+c


    =2\sqrt{x^2-4}-4\text{ arcsec}(\tfrac12x) + c

    This answer is correct, but there are other ways to write it. So, check it by differentiating, noting that:
    \frac{d}{dx}\big(\text{ arcsec}(\tfrac12x)\big) = \frac{2}{x\sqrt{x^2-4}}
    Grandad
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