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Math Help - triggy substitution

  1. #1
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    triggy substitution

    find the following integral:

    \int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx

    what i was thinking was:
    x = \sqrt{3}tan u
    dx = \sqrt{3}sec˛u du

    then i get:


    \int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra  c{5}{2}}} \cdot \sqrt{3} sec^2u du

    is this right?
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  2. #2
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    Quote Originally Posted by qzno View Post
    find the following integral:

    \int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx

    what i was thinking was:
    x = \sqrt{3}tan u
    dx = \sqrt{3}sec˛u du

    then i get:


    \int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra  c{5}{2}}} \cdot \sqrt{3} sec^2u du

    is this right?
    Yep - keep going!
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  3. #3
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    does that equal:

    \int^3_0 \frac{tan^3u}{\sqrt{3} sec^{\frac{5}{2}}u} du

    ????
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  4. #4
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    Quote Originally Posted by qzno View Post
    does that equal:

    \int^3_0 \frac{tan^3u}{\sqrt{3} sec^{\frac{5}{2}}u} du

    ????
    No! \int^{\pi/3}_0 \frac{\tan^3u \sec^2 u}{\sqrt{3} \sec^{5}u} du = \frac{1}{\sqrt{3}} \int^{\pi/3}_0 \frac{\tan^3u }{ \sec^{3}u} du Note - the limits change.
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  5. #5
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    why is it from 0 to pi/3 now?
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  6. #6
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    Quote Originally Posted by qzno View Post
    why is it from 0 to pi/3 now?
    You original limits were where x was equal to 3 and 0, you are now however working with u.

    Using your subsitution x = \sqrt{3}tan u, rearrange to get:

    u= arctan \frac{x}{\sqrt{3}}

    From this, sub in your two values of x, you then get the new values that danny mentioned.
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  7. #7
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    Quote Originally Posted by qzno View Post
    why is it from 0 to pi/3 now?
    The original problem was x = 0\; \text{to}\, x = 3. Introducing the substitution x = \sqrt{3} \tan u meanins that when you switch the problem in terms of u, the limits of integration will change. So, how do you find new limits - use the substitution. In your case

    x = 0\;\;\; \Rightarrow \;\;\; \sqrt{3} \tan u = 0\;\;\; \Rightarrow\; u = 0
    x = 3\;\;\; \Rightarrow \;\;\; \sqrt{3} \tan u = 3\;\;\; \Rightarrow\; u = \pi/3.
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  8. #8
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    ah i understand! : ) thanks!

    ill be posting my answer shortly hopefully
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  9. #9
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    i got my final answer to be:

    \frac{5}{24\sqrt{3}}
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  10. #10
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    Quote Originally Posted by qzno View Post
    i got my final answer to be:

    \frac{5}{24\sqrt{3}}
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  11. #11
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    Quote Originally Posted by qzno View Post
    find the following integral:

    \int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx

    what i was thinking was:
    x = \sqrt{3}tan u
    dx = \sqrt{3}sec˛u du

    then i get:


    \int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra  c{5}{2}}} \cdot \sqrt{3} sec^2u du

    is this right?
    alternatively you can put 3+x^2=u^2. then xdx=udu and so your integral becomes: \int_{\sqrt{3}}^{2 \sqrt{3}} \frac{u^2 - 3}{u^4} \ du and you know how to finish this!
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  12. #12
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    Hello,

    Another method, not necessarily more beautiful...
    \int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx

    Let t=\frac{1}{(3+x^2)^{3/2}}
    dt=\frac{dx \cdot 2x \cdot (-3/2)}{(3+x^2)^{5/2}}=\frac{-3x ~dx}{(3+x^2)^{5/2}}

    x=0 \Rightarrow t=3^{-3/2}=a
    x=3 \Rightarrow t=12^{-3/2}=b

    And it follows that 3+x^2=t^{-2/3} \Rightarrow x^2=t^{-2/3}-3

    Hence the integral is now :
    \int_a^b \frac{x \cdot x^2}{(3+x^2)^{5/2}} \cdot \frac{(3+x^2)^{5/2}}{-3x} ~dt=-\frac 13 \int_a^b x^2 ~dt=-\frac 13 \int_a^b t^{-2/3}-3 ~dt


    Okay...
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