+0  
 
0
62
3
avatar+55 

the integral of e^csc (π+t) • csc (π+t) cot (π+t) ? 

 May 2, 2020

Best Answer 

 #1
avatar+109520 
+3

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

 

 

 

 

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

 May 2, 2020
 #1
avatar+109520 
+3
Best Answer

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

 

 

 

 

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

Melody May 2, 2020
 #2
avatar+111329 
+1

Very nice solution, Melody   !!!!

 

 

cool cool cool

CPhill  May 2, 2020
 #3
avatar+109520 
+1

Thanks Chris :)

Melody  May 2, 2020

20 Online Users

avatar