Question 85524
f'(x) = e^(2x)sin(x)
~
[u = e^(2x)
[du = 2*e^(2x) dx
[dv = sin(x)
[v = - cos(x)
~
{{{int( e^(2x)sin(x), dx,a,b)}}}
{{{-e^(2x)cos(x) - int( -2*e^(2x)*cos(x) , dx,a,b)}}}
{{{-e^(2x)cos(x) + 2*int( e^(2x)*cos(x) , dx,a,b)}}}
~
[u = e^(2x)
[du = 2*e^(2x) dx
[dv = cos(x)
[v = sin(x)
~
{{{-e^(2x)cos(x) + 2*int( e^(2x)*cos(x) , dx,a,b)}}}
{{{-e^(2x)cos(x) + 2(e^(2x)sin(x) - int(2*e^(2x)*sin(x) , dx,a,b))}}}
{{{-e^(2x)cos(x) + 2(e^(2x)sin(x) - 2*int(e^(2x)*sin(x) , dx,a,b))}}}
{{{-e^(2x)cos(x) + 2e^(2x)sin(x) - 4*int(e^(2x)*sin(x) , dx,a,b)}}}
~
{{{y = int( e^(2x)sin(x), dx,a,b)}}}
~
{{{y = -e^(2x)cos(x) + 2e^(2x)sin(x) - 4*int(e^(2x)*sin(x) , dx,a,b)}}}
{{{y = -e^(2x)cos(x) + 2e^(2x)sin(x) - 4y}}}
{{{5y = e^(2x)(2*sin(x) - cos(x))}}}
{{{y = (1/5)e^(2x)(2*sin(x) - cos(x))}}}
~
Constant Rule:
~
{{{y = (1/5)e^(2x)(2*sin(x) - cos(x)) + C}}}