Question

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

Answer 1

`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`