# cot(x) and ln sin(x)

• May 17th 2012, 07:34 AM
vernal
cot(x) and ln sin(x)

Attachment 23891

thanks
• May 17th 2012, 07:41 AM
Sylvia104
Re: cot(x) and ln sin(x)
Try $\int_{\frac{\pi}2}^xf(t)\,\mathrm dt \leqslant \int_{\frac{\pi}2}^x\cot t\,\mathrm dt \leqslant \int_{\frac{\pi}2}^xg(t)\,\mathrm dt$ for some functions $f(t),g(t).$

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

PS: $f(t)=\frac{\mathrm d}{\mathrm dt}\left[t\cot t\right]$ and $g(t)\equiv0.$
• May 17th 2012, 08:41 AM
Sylvia104
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 $0\leqslant -\cot t \leqslant -f(t)$ for $\frac{\pi}2\leqslant t<\pi$ where $f(t) = \frac{\mathrm d}{\mathrm dt}\left[t\cot t\right] = \cot t-t\csc^2t.$

Then integrate with respect to $t$ from $t=\frac{\pi}2$ to $t=x$ and multiply through by $-1.$
• May 17th 2012, 11:28 AM
vernal
Re: cot(x) and ln sin(x)
Quote:

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

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

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

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

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