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Thread: cot(x) and ln sin(x)

  1. #1
    Member vernal's Avatar
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    cot(x) and ln sin(x)

    Hi, please help me, what can prove it

    cot(x)  and ln sin(x)-untitled.jpg

    thanks
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  2. #2
    Member Sylvia104's Avatar
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    Re: cot(x) and ln sin(x)

    Try $\displaystyle \int_{\frac{\pi}2}^xf(t)\,\mathrm dt \leqslant \int_{\frac{\pi}2}^x\cot t\,\mathrm dt \leqslant \int_{\frac{\pi}2}^xg(t)\,\mathrm dt$ for some functions $\displaystyle f(t),g(t).$

    By the way, you stated the range of $\displaystyle x$ as $\displaystyle x\leq\frac{\pi}2\leq\pi$ but I think you mean $\displaystyle \frac\pi2\leq x<\pi.$

    PS: $\displaystyle f(t)=\frac{\mathrm d}{\mathrm dt}\left[t\cot t\right]$ and $\displaystyle g(t)\equiv0.$
    Last edited by Sylvia104; May 17th 2012 at 07:51 AM.
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  3. #3
    Member Sylvia104's Avatar
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    Re: cot(x) and ln sin(x)

    I beg your pardon, the functions are negative in the given range so we need to do it this way:

    Show that $\displaystyle 0\leqslant -\cot t \leqslant -f(t)$ for $\displaystyle \frac{\pi}2\leqslant t<\pi$ where $\displaystyle f(t) = \frac{\mathrm d}{\mathrm dt}\left[t\cot t\right] = \cot t-t\csc^2t.$

    Then integrate with respect to $\displaystyle t$ from $\displaystyle t=\frac{\pi}2$ to $\displaystyle t=x$ and multiply through by $\displaystyle -1.$
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  4. #4
    Member vernal's Avatar
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    Re: cot(x) and ln sin(x)

    Quote Originally Posted by Sylvia104 View Post
    Try $\displaystyle \int_{\frac{\pi}2}^xf(t)\,\mathrm dt \leqslant \int_{\frac{\pi}2}^x\cot t\,\mathrm dt \leqslant \int_{\frac{\pi}2}^xg(t)\,\mathrm dt$ for some functions $\displaystyle f(t),g(t).$

    By the way, you stated the range of $\displaystyle x$ as $\displaystyle x\leq\frac{\pi}2\leq\pi$ but I think you mean $\displaystyle \frac\pi2\leq x<\pi.$

    PS: $\displaystyle f(t)=\frac{\mathrm d}{\mathrm dt}\left[t\cot t\right]$ and $\displaystyle g(t)\equiv0.$

    Sorry. yes .

    thanks
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  5. #5
    Member vernal's Avatar
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    Re: cot(x) and ln sin(x)

    thanks, but i can't understand. There another way?
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