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

thanks

2. ## 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 $\displaystyle g(t)\equiv0.$

3. ## 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 $\displaystyle -1.$

4. ## Re: cot(x) and ln sin(x)

Originally Posted by Sylvia104
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 . $\frac{\pi}{2}\leq x \leq \pi$

thanks

5. ## Re: cot(x) and ln sin(x)

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