How do you find the antiderivative of $e^(2x) * sin(4x)dx$ ?

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Answer 1 · 2018-04-08T17:57:53

$I=e^(2x)/10(sin4x-2cos4x)+c$
OR
$I=e^(2x)/sqrt(20)sin(4x-tan^-1 2)+c$

Here,

$I=inte^(2x)sin4xdx$

We know that,

$color(red)((1)inte^(ax) sinbxdx=e^(ax)/(a^2+b^2)(asinbx- bcosbx)+c$

Comparing we get, $a=2 and b=4$

So,

$I=e^(2x)/(2^2+4^2)(2sin4x-4cos4x)+c$

$I=e^(2x)/20(2sin4x-4cos4x)+c$

$I=e^(2x)/10(sin4x-2cos4x)+c$

OR

We also know that,

$color(red)((2)inte^(ax)sinbxdx=e^(ax)/sqrt(a^2+b^2)sin(bx- theta)+c$ ,

where, $theta=tan^-1(b/a)$

So,

$I=e^(2x)/sqrt(2^2+4^2)sin(4x-theta)+c$ ,

where, $theta=tan^-1(4/2)=tan^-1(2)$

Hence,

$I=e^(2x)/sqrt(20)sin(4x-tan^-1 2)+c$

How do you find the antiderivative of e^(2x) * sin(4x)dxe^(2x) * sin(4x)dx?