Melody

\(\displaystyle\int e^{csc (π+t)} * csc (π+t) cot (π+t)\; dt\\ =\displaystyle\int e^{-csc (t)} *-csc (t) cot (t)\; dt\\ =-\displaystyle\int e^{-csc (t)} *csc (t) cot (t)\; dt\\~\\ \qquad \;\;\;Let\;\;u=-csc(t)\\ \qquad \qquad \;\;u=-(sin(t))^{-1}\\ \qquad \qquad \;\;\frac{du}{dt}=(sin(t))^{-2}*cos(t)\\ \qquad \qquad \;\;\frac{du}{dt}=\frac{cos(t)}{(sin(t))^2}\\ \qquad \qquad \;\;\frac{du}{dt}=csc(t)cot(t)\\~\\ \qquad \qquad \;\;dt=\frac{du}{csc(t)cot(t)}\\~\\ =-\displaystyle\int e^{-csc (t)} *csc (t) cot (t)\; dt\\~\\ =-\displaystyle\int e^{u} *csc (t) cot (t)\; \frac{du}{csc(t)cot(t)}\\~\\ =-\displaystyle\int e^{u} du\\~\\ =-e^u+c\\ =-e^{-csc(t)}+c\)

LaTex:

\displaystyle\int e^{csc (π+t)} * csc (π+t) cot (π+t)\; dt\\
=\displaystyle\int e^{-csc (t)} *-csc (t)  cot (t)\; dt\\
=-\displaystyle\int e^{-csc (t)} *csc (t)  cot (t)\; dt\\~\\
\qquad \;\;\;Let\;\;u=-csc(t)\\
\qquad \qquad \;\;u=-(sin(t))^{-1}\\
\qquad \qquad \;\;\frac{du}{dt}=(sin(t))^{-2}*cos(t)\\
\qquad \qquad \;\;\frac{du}{dt}=\frac{cos(t)}{(sin(t))^2}\\
\qquad \qquad \;\;\frac{du}{dt}=csc(t)cot(t)\\~\\
\qquad \qquad \;\;dt=\frac{du}{csc(t)cot(t)}\\~\\

=-\displaystyle\int e^{-csc (t)} *csc (t)  cot (t)\; dt\\~\\
=-\displaystyle\int e^{u} *csc (t)  cot (t)\; \frac{du}{csc(t)cot(t)}\\~\\
=-\displaystyle\int e^{u} du\\~\\
=-e^u+c\\
=-e^{-csc(t)}+c

Thanks LuckyDuck,

Choy, if you want to repost please give the address of your original question and ask people to answer there. 

Hi Littlemixfan,

Find the number of pairs of integers (x,y) with x is less than 0, y is less than 10, that satisfy

\(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\)

First I want to look at formula restrictions.   

\(y\ne0\\ x\ne0\\ x\ne10\)

We also have the question restriction of

\(x<0\qquad and \qquad y<10\)

So where can it be so far:

NOW

\(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\\ \frac{1}{\frac{x-10}{x}} > 1 - \frac{5}{y}\\ \frac{x}{x-10} > \frac{x-10}{x-10} - \frac{5}{y}\\ \frac{x-(x-10)}{x-10} > - \frac{5}{y}\\ \frac{10}{x-10} > - \frac{5}{y}\\ \frac{x-10} {10}< - \frac{y}{5}\\ - \frac{y}{5}>\frac{x-10} {10} \\ \frac{y}{5}<\frac{-x+10} {10} \\ \frac{y}{1}<\frac{-x+10} {2} \\ y<-0.5x+5 \)

So that is underneath the line y=-0.5x+5

NOTE: Maybe I made an error but I do not think so. Maybe it was meant to be a   <   sign.

Also,:

I left the > sign in for all my working but it would have been easier just to solve for = and then test the regions afterwards.

So far I have these 2 regions.  But since there are 2 regions one where y<0 and one where y>0 I will test both.

Test  (-1,+1) and (-1,-1)

\(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\\test\;(-1,1)\\ \frac{1}{11}>-4\quad true \\ test\;(-1,-1)\\ \frac{1}{11}>6\quad false \\ \)

So here is the region:

I think there is an infinite number of integer pair solutions.

LaTex:

\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\\
\frac{1}{\frac{x-10}{x}} > 1 - \frac{5}{y}\\
\frac{x}{x-10} > \frac{x-10}{x-10} - \frac{5}{y}\\
\frac{x-(x-10)}{x-10} >  - \frac{5}{y}\\
\frac{10}{x-10} >  - \frac{5}{y}\\
\frac{x-10} {10}<  - \frac{y}{5}\\
- \frac{y}{5}>\frac{x-10} {10} \\

\frac{y}{5}<\frac{-x+10} {10} \\

\frac{y}{1}<\frac{-x+10} {2} \\

y<-0.5x+5

100*10 is only 1000 ants.  There are a lot more ants than that!  So you have an error in your question.

If your question was correct.

You'd have

     18*10*100*10 g   divided by  4000kg

=    180000g             divided by   4000kg

=        180Kg             divided by   4000Kg

Note that the ants weigh less than one hippo

=        18/400     

=  0.045

=   4.5*10^(-2)     

Thanks very much Alan.  

Thanks to admin for the fix too.   

Why have you not answered my question?

I suppose it was because you just reposted and another answerer spoonfed you with the answer.

Yes I saw!

Well, you at least posted this pic the right way around. That is a small step forward.

Stop reposting!  I was already answered this with you! 

You will be banned from the forum if you do not start behaving appropriately.

I expect you are smart and better still you are interested and conscientious.

Those three traits can take you a long way.