Question

How do you find the antiderivative of `(e^(2x))sin3x`?

Answer 1

`int \ e^(2x)sin3x \ dx = e^(2x)/13(2sin3x -3cos3x ) + C`

Explanation:

We could use a traditional double application of Integration By Parts. Here is a slightly different approach.

Let:

`s = e^(2x)sin3x \ \ \ \` and `I_s = int e^(2x)sin3x`
`c = e^(2x)cos3x \ \ \ \` and `I_c = int e^(2x)cos3x`

Differentiating wrt `x` we get:

`(ds)/dx = e^(2x)(d/dx sin3x) + (d/dx e^(2x))sin3x`
`\ \ \ \ \ \ = 3e^(2x)cos3x + 2e^(2x)sin3x`

`(dc)/dx = e^(2x)(d/dx cos3x) + (d/dx e^(2x))cos3x`
`\ \ \ \ \ \ = -3e^(2x)sin3x + 2e^(2x)cos3x`

Now integrate the above results:

`int \ (ds)/dx \ dx = int \ 3e^(2x)cos3x + 2e^(2x)sin3x \ dx`
`=> s= 3I_c + 2I_s` ... [A]

`int \ (dc)/dx \ dx = int \ -3e^(2x)sin3x + 2e^(2x)cos3x \ dx`
`=> c = -3I_s + 2I_c` ... [B]

3Eq [A] + 2Eq [B}:

`3s+2c = 9I_c + 6I_s -6I_s + 4I_c`
`:. 3s+2c = 13I_c`
`:. I_c = 1/13(3s+2c)`

From [A] we also get:

`s = 3/13(3s+2c) + 2I_s`
`:. s = 9/13s+6/13c + 2I_s`
`:. I_s = 1/13(2s -3c)`

Hence we get the two results:

`int \ e^(2x)cos3x \ dx = 1/13(3e^(2x)sin3x+2e^(2x)cos3x) + C`
`" " = e^(2x)/13(3sin3x+2cos3x) + C`

`int \ e^(2x)sin3x \ dx = 1/13(2e^(2x)sin3x -3e^(2x)cos3x ) + C`
`" " = e^(2x)/13(2sin3x -3cos3x ) + C`