Question

integrate sinhx/cosh^2 x dx,
do i have to do substitution here

Answer 1

Yes.

Answer 2

The most natural sub is either going to be u=sinh(x) or u=cosh(x).
Any ideas which would be better? :)

Answer 3

I guess I should say,
Any ideas which sub `will actually work`? :)

Answer 4

u=coshx?

Answer 5

\[\large\rm \int\limits\frac{\sinh x}{\cosh^2x}dx\qquad=\qquad \int\limits \frac{1}{\color{royalblue}{\cosh}^2\color{royalblue}{x}}\left(\color{orangered}{\sinh x~dx}\right)\]Good :)

Answer 6

\[\large\rm \color{royalblue}{u=\cosh x},\qquad\qquad du=?\]

Answer 7

sinhx

Answer 8

\[\large\rm \color{royalblue}{u=\cosh x},\qquad\qquad \color{orangered}{du=\sinh x~dx}\]Cool

Answer 9

so it will leave \[\int\limits \frac{ 1 }{ coshx }\]

Answer 10

Woops, replace all of the colors,\[\large\rm \int\limits\limits \frac{1}{\color{royalblue}{\cosh}^2\color{royalblue}{x}}\left(\color{orangered}{\sinh x~dx}\right)\quad=\quad \int\limits\limits \frac{1}{\color{royalblue}{u}^2}\left(\color{orangered}{du}\right)\]Understand?

Answer 11

yup

Answer 12

And then just power rule through your u, ya?\[\large\rm =\int\limits u^{-2}du\]

Answer 13

ok, thanks