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Thread: integral

  1. #1
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    integral

    Let $\displaystyle I_n$ be

    $\displaystyle \int \frac{dx}{(a+\sin{x})^n}$

    Find the relations between $\displaystyle I_n , I_{n+1} $ and $\displaystyle I_{n+2}$

    Also find $\displaystyle I_1 $ and$\displaystyle I_2$
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by simplependulum View Post
    Let $\displaystyle I_n$ be

    $\displaystyle \int \frac{dx}{(a+\sin{x})^n}$

    Find the relations between $\displaystyle I_n , I_{n+1} $ and $\displaystyle I_{n+2}$

    Also find $\displaystyle I_1 $ and$\displaystyle I_2$
    $\displaystyle {I_n} = \int {\frac{{dx}}{{{{\left( {a + \sin x} \right)}^n}}}} = \int {\frac{{{{\left( {a + \sin x} \right)}^2}}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}} dx =$

    $\displaystyle = \int {\frac{{{a^2} + 2a\sin x + {{\sin }^2}x}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}} dx = \int {\frac{{{a^2} + 1 + 2a\sin x - {{\cos }^2}x}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}} dx =$

    $\displaystyle = \left( {{a^2} + 1} \right)\underbrace {\int {\frac{{dx}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}} }_{{I_{n + 2}}} + 2a\int {\frac{{\sin x}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}dx} - \int {\frac{{{{\cos }^2}x}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}dx} .$

    $\displaystyle \int {\frac{{\sin x}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}dx} = \int {\frac{{a + \sin x - a}}{{{{\left( {a + \sin x} \right)}^{n +2}}}}dx} = \underbrace {\int {\frac{{dx}}{{{{\left( {a + \sin x} \right)}^{n + 1}}}}} }_{{I_{n + 1}}} - a\underbrace {\int {\frac{{dx}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}} }_{{I_{n + 2}}}.$

    $\displaystyle \int {\frac{{{{\cos }^2}x}}{{{{\left( {a + \sin x} \right)}^{n + 2}}}}dx} = - \frac{1}{{n + 1}}\int {\cos xd\left( {\frac{1}{{{{\left( {a + \sin x} \right)}^{n + 1}}}}} \right)} =$

    $\displaystyle = - \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} - \frac{1}{{n + 1}}\int {\frac{{\sin x}}{{{{\left( {a + \sin x} \right)}^{n + 1}}}}dx} =$

    $\displaystyle = - \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} - \frac{1}{{n + 1}}\int {\frac{{a + \sin x - a}}{{{{\left( {a + \sin x} \right)}^{n + 1}}}}dx} =$

    $\displaystyle = - \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} - \frac{1}{{n + 1}}\underbrace {\int {\frac{{dx}}{{{{\left( {a + \sin x} \right)}^n}}}} }_{{I_n}} + \frac{a}{{n + 1}}\underbrace {\int {\frac{{dx}}{{{{\left( {a + \sin x} \right)}^{n + 1}}}}} }_{{I_{n + 1}}}.$

    So we have

    $\displaystyle {I_n} = \left( {{a^2} + 1} \right){I_{n + 2}} + 2a\left( {{I_{n + 1}} - a{I_{n + 2}}} \right) -$$\displaystyle \left( { - \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} - \frac{1}
    {{n + 1}}{I_n} + \frac{a}{{n + 1}}{I_{n + 1}}} \right) \Leftrightarrow $

    $\displaystyle \Leftrightarrow \frac{n}{{n + 1}}{I_n} = \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} + \frac{{a\left( {2n + 3} \right)}}{{n + 1}}{I_{n + 1}} + \left( {1 - {a^2}} \right){I_{n + 2}} \Leftrightarrow$

    $\displaystyle \Leftrightarrow {I_n} = \frac{{\cos x}}{{n{{\left( {a + \sin x} \right)}^{n + 1}}}} + \frac{{a\left( {2n + 3} \right)}}{n}{I_{n + 1}} + \frac{{\left( {n + 1} \right)\left( {1 - {a^2}} \right)}}{n}{I_{n + 2}}.$
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by simplependulum View Post
    Let $\displaystyle I_n$ be

    $\displaystyle \int \frac{dx}{(a+\sin{x})^n}$

    Find the relations between $\displaystyle I_n , I_{n+1} $ and $\displaystyle I_{n+2}$

    Also find $\displaystyle I_1 $ and$\displaystyle I_2$
    For $\displaystyle I_1$

    $\displaystyle \int {\frac{{dx}}{{a + \sin x}}} =$$\displaystyle \int {\frac{{dx}}
    {{a + \frac{{2\tan \left( {{x \mathord{\left/
    {\vphantom {x 2}} \right.
    \kern-\nulldelimiterspace} 2}} \right)}}
    {{1 + {{\tan }^2}\left( {{x \mathord{\left/
    {\vphantom {x 2}} \right.
    \kern-\nulldelimiterspace} 2}} \right)}}}}} = \left\{ \begin{gathered}
    \tan \frac{x}
    {2} = t, \hfill \\
    x = 2\arctan t, \hfill \\
    dx = \frac{2}
    {{1 + {t^2}}}dt \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = 2\int {\frac{1}{{a + \frac{{2t}}{{1 + {t^2}}}}}\frac{{dt}}
    {{1 + {t^2}}}} = 2\int {\frac{{dt}}{{a{t^2} + 2t + a}}} = 2a\int {\frac{{dt}}{{{a^2}{t^2} + 2at + 1 + {a^2} - 1}}} =$

    $\displaystyle = 2a\int {\frac{{dt}}
    {{{{\left( {at + 1} \right)}^2} + {a^2} - 1}}} = \frac{{2a}}
    {{{a^2} - 1}}\int {\frac{{dt}}
    {{{{\left( {{{\left( {at + 1} \right)} \mathord{\left/
    {\vphantom {{\left( {at + 1} \right)} {\sqrt {{a^2} - 1} }}} \right.
    \kern-\nulldelimiterspace} {\sqrt {{a^2} - 1} }}} \right)}^2} + 1}}} =$

    $\displaystyle = \left\{ \begin{gathered}
    \frac{{at + 1}}
    {{\sqrt {{a^2} - 1} }} = u, \hfill \\
    \frac{a}
    {{\sqrt {{a^2} - 1} }}dt = du \hfill \\
    \end{gathered} \right\} = \frac{2}
    {{\sqrt {{a^2} - 1} }}\int {\frac{{du}}
    {{{u^2} + 1}}} = \frac{2}
    {{\sqrt {{a^2} - 1} }}\arctan u + C =$

    $\displaystyle = \frac{2}{{\sqrt {{a^2} - 1} }}\arctan \frac{{at + 1}}
    {{\sqrt {{a^2} - 1} }} + C = \frac{2}{{\sqrt {{a^2} - 1} }}\arctan \frac{{a\tan \frac{x}{2} + 1}}{{\sqrt {{a^2} - 1} }} + C.$
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