How do you integrate $sin(x)*(e)^(2 x) dx$ ?

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Answer 1 · 2016-07-01T16:02:30

Solution: $int e^(2x)sin(x)dx = 2/5e^(2x)sin(x) - 1/5e^(2x)cos(x) + C$

To obtain the formula for IBP, we start with the product rule:

$(u*v)' = u'*v + u*v'$ where u and v are functions of x in this case

Rearranging:

$u*v' = (u*v)' - u'*v$

Integrating:

$int u*v' dx = u*v - int u'*v dx$

For this integral, we are going to set $u(x) = sin(x)$ and $v'(x) = e^(2x)$

$u'(x) = cos(x)$ and $v(x) = 1/2e^(2x)$

So, $int e^(2x)sinxdx = 1/2e^(2x)sin(x) - 1/2*int cos(x)e^(2x) dx$

Now do IBP again on the second term:

$u(x) = cos(x)$ and $v'(x) = e^(2x)$
$u'(x) = -sin(x)$ and $v(x) = 1/2e^(2x)$

$int e^(2x)sin(x)dx = 1/2e^(2x)sin(x) - 1/4e^(2x)cos(x) - 1/4*int e^(2x)sin(x)dx$

As you can see, the integral on the LHS is the same as the integral on the RHS so we collect like terms and obtain:

$5/4*int e^(2x)sin(x)dx = 1/2e^(2x)sin(x) - 1/4e^(2x)cos(x)$

$int e^(2x)sin(x)dx = 2/5e^(2x)sin(x) - 1/5e^(2x)cos(x) + C$

How do you integrate sin(x)*(e)^(2 x) dxsin(x)*(e)^(2 x) dx?