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Math Help - integration

  1. #1
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    integration

    1) integrate sin3x cos7x

    i assumed u=3x and another time u=7x
    then i came up with the ans:
    -1/3 cos3x * 1/7 sin7x

    2) integrate sqrrt (1-4x^2) dx
    using the integration substitution to get the form of sqrt(a^2-x^2)
    let a^2 = 1/4 x=1/2sinx
    sqrt(1/4-1/4sin^2x)
    but then i couldnt continue...
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  2. #2
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    1) \int \sin 3x \cos 7x~dx
    No way! You can't split an integral in two and integrate and then multiply them.. You have to integrate \int \sin 3x \cos 7x~dx as a whole.

    The best way is first using the trig identity \cos a \sin b = \frac{1}{2}\cdot [\sin(a+b)-\sin(a-b)], then integrating.

    Remember: \int f(x)g(x)~dx is NOT \left( \int f(x)~dx \right ) \cdot \left( \int g(x)~dx\right )

    But \int f(x)+g(x)~dx is equal to \int f(x)~dx+\int g(x)~dx


    -----------------------



    2) \int \sqrt{1-4x^2}~dx

    Can you see that this is equal to \int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx

    Now you can use u-substitution and then integrate it by parts.
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  3. #3
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    Quote Originally Posted by wutever View Post
    1) integrate sin3x cos7x

    i assumed u=3x and another time u=7x
    then i came up with the ans:
    -1/3 cos3x * 1/7 sin7x

    Mr F says: Uh uh. You need to manipulate the compound angle formulae to get the following identity:

    {\color{red} \sin A \cos B = \frac{\sin (A + B) + \sin (A - B)}{2}}.

    So {\color{red} \sin (3x) \cos (7x) = \frac{\sin (10x) + \sin (-4x)}{2} = \frac{\sin (10x) - \sin (4x)}{2}}.

    2) integrate sqrrt (1-4x^2) dx
    using the integration substitution to get the form of sqrt(a^2-x^2)
    let a^2 = 1/4 x=1/2sinx
    sqrt(1/4-1/4sin^2x)
    but then i couldnt continue...

    Mr F says: Perhaps if I write {\color{red} \frac{1}{4} - \frac{1}{4} \sin^2 x = \frac{1}{4} (1 - \sin^2 x)} you'll see how the Pythagorean Identity makes life simple ......
    ..
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  4. #4
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    ok i get the first one but i cant fully understand the second one !!!!
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  5. #5
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    Quote Originally Posted by wutever View Post
    ok i get the first one but i cant fully understand the second one !!!!


    1 - \sin^2 x = \cos^2 x .......
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  6. #6
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    okk i know that but i have a weired ans :
    1/4 sin inverse (2x) + 1/2 x sqrt (1-4x^2)

    so how ??
    and why <br /> <br />
\int \sqrt{1-4x^2}~dx<br />

    is equal to

    <br /> <br />
\int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx<br />
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  7. #7
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    Quote Originally Posted by wutever View Post
    okk i know that but i have a weired ans :
    1/4 sin inverse (2x) + 1/2 x sqrt (1-4x^2)

    Mr F says: *Ahem* .... + C

    so how ??

    Mr F says: The answer you got is correct. If you differentiated it, you would see that you get \sqrt{1 - 4x^2} .......

    and why <br /> <br />
\int \sqrt{1-4x^2}~dx<br />

    is equal to

    <br /> <br />
\int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx<br />

    Mr F says: Basic algebra. Multiply {\color{red}\sqrt{1 - 4x^2}} by {\color{red}1 = \frac{\sqrt{1 - 4x^2}}{\sqrt{1 - 4x^2}}}
    ..
    Last edited by mr fantastic; June 1st 2008 at 01:55 PM. Reason: Fixed the color.
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  8. #8
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    For the second integral you want to use the substitution:

    2u = sin(x)

    You then get: Sqrt (1 - (sin(x))^2)

    Which = sqrt ( (cos(x))^2 )
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