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Math Help - integral

  1. #1
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    integral

    Let  I_n be

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

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

    Also find I_1 and  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  I_n be

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

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

    Also find I_1 and  I_2
    {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 =

    = \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 =

    = \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} .

    \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}}}.

    \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)}  =

    =  - \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}  =

    =  - \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}  =

    =  - \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

    {I_n} = \left( {{a^2} + 1} \right){I_{n + 2}} + 2a\left( {{I_{n + 1}} - a{I_{n + 2}}} \right) - \left( { - \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} - \frac{1}<br />
{{n + 1}}{I_n} + \frac{a}{{n + 1}}{I_{n + 1}}} \right) \Leftrightarrow

    \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

    \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  I_n be

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

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

    Also find I_1 and  I_2
    For I_1

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

    = 2\int {\frac{1}{{a + \frac{{2t}}{{1 + {t^2}}}}}\frac{{dt}}<br />
{{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}}}  =

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

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

    = \frac{2}{{\sqrt {{a^2} - 1} }}\arctan \frac{{at + 1}}<br />
{{\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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