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

1 Answer
Apr 8, 2018

Answer:

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

Explanation:

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#