1. ## 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$\displaystyle I_2$

2. Originally Posted by simplependulum
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) -$$\displaystyle \left( { - \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} - \frac{1} {{n + 1}}{I_n} + \frac{a}{{n + 1}}{I_{n + 1}}} \right) \Leftrightarrow \displaystyle \Leftrightarrow \frac{n}{{n + 1}}{I_n} = \frac{{\cos x}}{{\left( {n + 1} \right){{\left( {a + \sin x} \right)}^{n + 1}}}} + \frac{{a\left( {2n + 3} \right)}}{{n + 1}}{I_{n + 1}} + \left( {1 - {a^2}} \right){I_{n + 2}} \Leftrightarrow \displaystyle \Leftrightarrow {I_n} = \frac{{\cos x}}{{n{{\left( {a + \sin x} \right)}^{n + 1}}}} + \frac{{a\left( {2n + 3} \right)}}{n}{I_{n + 1}} + \frac{{\left( {n + 1} \right)\left( {1 - {a^2}} \right)}}{n}{I_{n + 2}}. 3. Originally Posted by simplependulum Let \displaystyle I_n be \displaystyle \int \frac{dx}{(a+\sin{x})^n} Find the relations between \displaystyle I_n , I_{n+1} and \displaystyle I_{n+2} Also find \displaystyle I_1 and\displaystyle I_2 For \displaystyle I_1 \displaystyle \int {\frac{{dx}}{{a + \sin x}}} =$$\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\} =$

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

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

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

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