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Thread: Integrals Rp_4-5

  1. #1
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    Integrals Rp_4-5

    $\displaystyle \int_1^2 \frac{1}{(x^2-2x+5)^{3/2}}dx= $

    $\displaystyle \int 5^{\sqrt{7x+2}}dx= $

    In the secound problem i tried maybe $\displaystyle e^{(\sqrt{7x+2}*ln{5})}$ but I don't know how to continue...
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by gilyos View Post
    $\displaystyle \int_1^2 \frac{1}{(x^2-2x+5)^{3/2}}dx= $

    $\displaystyle \int {\frac{{dx}}
    {{{{\left( {{x^2} - 2x + 5} \right)}^{3/2}}}}} = \int {\frac{{dx}}
    {{{{\left( {{{\left( {x - 1} \right)}^2} + 4} \right)}^{3/2}}}}} = \left\{ \begin{gathered}
    x - 1 = 2\tan u, \hfill \\
    dx = \frac{{2du}}
    {{{{\cos }^2}u}} \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = \frac{1}
    {4}\int {\frac{{du}}
    {{{{\cos }^2}u{{\left( {{{\tan }^2}u + 1} \right)}^{3/2}}}}} = \frac{1}
    {4}\int {\frac{{du}}
    {{{{\cos }^2}u{{\left( {\frac{1}
    {{{{\cos }^2}u}}} \right)}^{3/2}}}}} = \frac{1}
    {4}\int {\frac{{{{\cos }^3}u}}
    {{{{\cos }^2}u}}\,du} =$

    $\displaystyle = \frac{1}
    {4}\int {\cos u\,du} = \frac{1}
    {4}\sin u + C = \frac{1}
    {4}\sin \arctan \frac{{x - 1}}
    {2} + C =$

    $\displaystyle = \frac{1}
    {4}\frac{{\frac{{x - 1}}
    {2}}}
    {{\sqrt {1 + {{\left( {\frac{{x - 1}}
    {2}} \right)}^2}} }} + C = \frac{{x - 1}}
    {{4\sqrt {{x^2} - 2x + 5} }} + C.$

    Finally

    $\displaystyle \int\limits_1^2 {\frac{{dx}}
    {{{{\left( {{x^2} - 2x + 5} \right)}^{3/2}}}}} = \left. {\frac{{x - 1}}
    {{4\sqrt {{x^2} - 2x + 5} }}} \right|_1^2 = \frac{1}
    {{4\sqrt 5 }} = \frac{{\sqrt 5 }}
    {{20}}.$
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by gilyos View Post
    $\displaystyle \int 5^{\sqrt{7x+2}}dx= $

    In the secound problem i tried maybe $\displaystyle e^{(\sqrt{7x+2}*ln{5})}$ but I don't know how to continue...
    Try this substitution $\displaystyle 7x + 2 = u^2$, after the integration by parts:

    $\displaystyle \int {{5^{\sqrt {7x + 2} }}\,dx} = \left\{ \begin{gathered}
    7x + 2 = {u^2} \hfill \\
    dx = \frac{{2u}}
    {7}\,du \hfill \\
    \end{gathered} \right\} = \frac{2}
    {7}\int {{5^u}u\,du} = \frac{{2 \cdot {5^u}u}}
    {{7\ln 5}} - \frac{2}
    {{7\ln 5}}\int {{5^u}\,du} =$

    $\displaystyle = \frac{{2 \cdot {5^u}u}}
    {{7\ln 5}} - \frac{{2 \cdot {5^u}}}
    {{7{{\ln }^2}5}} + C = \left( {\frac{{2u}}
    {{7\ln 5}} - \frac{2}
    {{7{{\ln }^2}5}}} \right){5^u} + C =$

    $\displaystyle = \left( {\frac{{2\sqrt {7x + 2} }}
    {{7\ln 5}} - \frac{2}
    {{7{{\ln }^2}5}}} \right){5^{\sqrt {7x + 2} }} + C.$
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    $\displaystyle = \frac{1}
    {4}\int {\cos u\,du} = \frac{1}
    {4}\sin u + C = \frac{1}
    {4}\sin \arctan \frac{{x - 1}}
    {2} + C =$



    How you transposed this ? from sin(arctan)

    to this :

    $\displaystyle = \frac{1}
    {4}\frac{{\frac{{x - 1}}
    {2}}}
    {{\sqrt {1 + {{\left( {\frac{{x - 1}}
    {2}} \right)}^2}} }} + C = \frac{{x - 1}}
    {{4\sqrt {{x^2} - 2x + 5} }} + C.$
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  5. #5
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    Quote Originally Posted by gilyos View Post
    $\displaystyle = \frac{1}
    {4}\int {\cos u\,du} = \frac{1}
    {4}\sin u + C = \frac{1}
    {4}\sin \arctan \frac{{x - 1}}
    {2} + C =$



    How you transposed this ? from sin(arctan)

    to this :

    $\displaystyle = \frac{1}
    {4}\frac{{\frac{{x - 1}}
    {2}}}
    {{\sqrt {1 + {{\left( {\frac{{x - 1}}
    {2}} \right)}^2}} }} + C = \frac{{x - 1}}
    {{4\sqrt {{x^2} - 2x + 5} }} + C.$
    $\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by gilyos View Post
    $\displaystyle = \frac{1}
    {4}\int {\cos u\,du} = \frac{1}
    {4}\sin u + C = \frac{1}
    {4}\sin \arctan \frac{{x - 1}}
    {2} + C =$



    How you transposed this ? from sin(arctan)

    to this :

    $\displaystyle = \frac{1}
    {4}\frac{{\frac{{x - 1}}
    {2}}}
    {{\sqrt {1 + {{\left( {\frac{{x - 1}}
    {2}} \right)}^2}} }} + C = \frac{{x - 1}}
    {{4\sqrt {{x^2} - 2x + 5} }} + C.$
    Treat $\displaystyle \vartheta=\arctan\frac{x-1}{2}$. Construct a triangle from this and then find $\displaystyle \sin\vartheta$.

    You should then see that $\displaystyle \sin\arctan\frac{x-1}{2}=\frac{x-1}{\sqrt{(x-1)^2+4}}=\frac{x-1}{\sqrt{x^2-2x+5}}$ as DeMath stated.
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  7. #7
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    Quote Originally Posted by Chris L T521 View Post
    Treat $\displaystyle \vartheta=\arctan\frac{x-1}{2}$. Construct a triangle from this and then find $\displaystyle \sin\vartheta$.

    You should then see that $\displaystyle \sin\arctan\frac{x-1}{2}=\frac{x-1}{\sqrt{(x-1)^2+4}}=\frac{x-1}{\sqrt{x^2-2x+5}}$ as DeMath stated.

    So , if i get it right

    $\displaystyle arcsin\frac{1}{2} = \frac{1}{\sqrt{1^2+2^2}} - \frac{1}{\sqrt{5}}$

    ?
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  8. #8
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    Quote Originally Posted by gilyos View Post
    So , if i get it right

    $\displaystyle arcsin\frac{1}{2} = \frac{1}{\sqrt{1^2+2^2}} - \frac{1}{\sqrt{5}}$

    ?
    See my post:

    $\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$.

    So if $\displaystyle \theta = \arctan{\frac{x - 1}{2}}$ then

    $\displaystyle \sin{\arctan{\frac{x - 1}{2}}} = \frac{\tan{\arctan{\frac{x - 1}{2}}}}{\sqrt{1 + \left(\tan{\arctan{\frac{x - 1}{2}}}\right)^2}}$

    $\displaystyle = \frac{\frac{x - 1}{2}}{\sqrt{1 + \left(\frac{x - 1}{2}\right)^2}}$

    $\displaystyle = \frac{\frac{x - 1}{2}}{\sqrt{\frac{4 + (x - 1)^2}{4}}}$

    $\displaystyle = \frac{\frac{x - 1}{2}}{\frac{\sqrt{4 + x^2 - 2x + 1}}{2}}$

    $\displaystyle = \frac{x - 1}{\sqrt{x^2 - 2x + 5}}$.
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