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Thread: Another integration problem

  1. #1
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    Another integration problem

    If $\displaystyle \displaystyle I_n=\int\frac{\cos^{2n} x}{\sin x} dx$, write down a similar expression for $\displaystyle I_{n+1}$. Hence, by using the identity $\displaystyle \cos^2 x+\sin^2 x\equiv 1$, prove that $\displaystyle (2n+1)I_{n+1}=(2n+1)I_n+\cos^{2n+1} x$. Hence or otherwise determine $\displaystyle \displaystyle\int\frac{\cos^6 x}{\sin x} dx$

    I found $\displaystyle I_{n+1}=\frac{\cos^{2n+2}}{\sin x} dx$
    But I don't know how to do the next part.
    Thanks!
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  2. #2
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    $\displaystyle \displaystyle{I_n = \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx}}$.

    Then $\displaystyle \displaystyle{I_{n+1} = \int{\frac{\cos^{2(n+1)}{x}}{\sin{x}}\,dx}}$

    $\displaystyle \displaystyle{= \int{\frac{\cos^{2n+2}{x}}{\sin{x}}\,dx}}$

    $\displaystyle \displaystyle{= \int{\cos^2{x}\,\frac{\cos^{2n}{x}}{\sin{x}}\,dx}}$

    $\displaystyle \displaystyle{= \int{(1 - \sin^2{x})\,\frac{\cos^{2n}{x}}{\sin{x}}\,dx}}$

    $\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}} - \sin{x}\cos^{2n}{x}\,dx}}$

    $\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \int{\sin{x}\cos^{2n}{x}\,dx}}$

    $\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \int{u^{2n}\,du}}$ after making the substitution $\displaystyle \displaystyle{u = \cos{x}}$

    $\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \frac{u^{2n + 1}}{2n + 1}}$

    $\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \frac{\cos^{2n + 1}{x}}{2n+1}}$

    $\displaystyle \displaystyle{= I_n - \frac{\cos^{2n + 1}{x}}{2n + 1}}$.


    Since we have

    $\displaystyle \displaystyle{I_{n + 1} = I_n - \frac{\cos^{2n + 1}{x}}{2n + 1}}$

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

    $\displaystyle \displaystyle{(2n + 1)\,I_{n + 1} = (2n + 1)\,I_n - \cos^{2n + 1}{x}}$.


    Now since you're trying to find $\displaystyle \displaystyle{\int{\frac{\cos^6{x}}{\sin{x}}\,dx}}$, this is $\displaystyle I_6$.

    Then you have

    $\displaystyle \displaystyle{I_6 = I_5 - \frac{\cos^{11}{x}}{11}}$.

    This means you will need to find $\displaystyle I_5, I_4, I_3, I_2$ and $\displaystyle I_1$.
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