Question 637388
<pre>
{{{3e^x*tan(y)dx}}} + {{{(1-e^x)*sec^2(y)dy}}} = 0

Separate the variables by dividing through by {{{(1-e^x)tan(y)
}}}

{{{(3e^x*tan(y)dx)/((1-e^x)tan(y))}}} + {{{((1-e^x)*sec^2(y)dy)/((1-e^x)tan(y))}}} = {{{0/((1-e^x)tan(y))}}}

{{{(3e^x*cross(tan(y))dx)/((1-e^x)cross(tan(y)))}}} + {{{((cross(1-e^x))*sec^2(y)dy)/((cross(1-e^x))tan(y))}}} = 0

{{{(3e^x*dx)/(1-e^x)}}} + {{{(sec^2(y)dy)/tan(y)}}} = 0

Now we integrate all three terms:

{{{int((3e^x)dx/(1-e^x)))}}} + {{{int((sec^2(y)dy)/tan(y))}}}= {{{int(0)}}}

Both terms on the left can be integrated using the formula {{{int(du/u)}}} = ln|u|+C
The second integral is already set up for that. We take the 3 out
of the integral on the first term:

{{{3int((e^x)dx/(1-e^x)))}}} + {{{int((sec^2(y)dy)/tan(y))}}}= {{{int(0)}}}

We need a negative sign in the numerator of the first fraction so it will
be in the form:

{{{-3int((-e^x)dx/(1-e^x)))}}} + {{{int((sec^2(y)dy)/tan(y))}}}= {{{int(0)}}}

Integrating using ln(C) for the arbitrary constant so everything will
be natural logs:

-3·ln|1-e<sup>x</sup>| + ln|tan(y)| = ln(C)

               ln|tan(y)| = 3·ln|1-e<sup>x</sup>| + ln(C)

Use a rule of logs to move the 3 to an exponent:

               ln|tan(y)| = ln|1-e<sup>x</sup>|<sup>3</sup> + ln(C)

Write the right side as the natural log of a product:

               ln|tan(y)| = ln[C|1-e<sup>x</sup>|<sup>3</sup>]

Take anti-logs of both sides:

               |tan(y)| = C|1-e<sup>x</sup>|<sup>3</sup>

Since the constant C can be positive or negative, we don't need
the absolute values:

               tan(y) = C(1-e<sup>x</sup>)<sup>3</sup>

And if we like we can solve for y by taking arctangents of both sides:

                    y = arctan[C(1-e<sup>x</sup>)<sup>3</sup>]

Edwin</pre>