Hi, please help me, what can prove it

Attachment 23891

thanks

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- May 17th 2012, 07:34 AMvernalcot(x) and ln sin(x)
Hi, please help me, what can prove it

Attachment 23891

thanks - May 17th 2012, 07:41 AMSylvia104Re: 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 $\displaystyle g(t)\equiv0.$ - May 17th 2012, 08:41 AMSylvia104Re: 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 $\displaystyle -1.$ - May 17th 2012, 11:28 AMvernalRe: cot(x) and ln sin(x)

Sorry. yes . http://latex.codecogs.com/gif.latex?...x%20\leq%20\pi

thanks - May 17th 2012, 11:34 AMvernalRe: cot(x) and ln sin(x)
thanks, but i can't understand.:( There another way?