Question

What is the integral of `int (e^x * cos 2x) dx`?

Answer 1

`int \ e^(x)cos2x \ dx = e^(x)/5(2sin2x+cos2x) + C`

Explanation:

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

Let:

`s = e^(x)sin2x \ \ \ \ ` and `I_s = int e^(x)sin2x`
`c = e^(x)cos2x \ \ \ \ ` and `I_c = int e^(x)cos2x`

Differentiating wrt `x` we get:

`(ds)/dx = e^(x)(d/dx sin2x) + (d/dx e^(x))sin2x`
`\ \ \ \ \ \ = 2e^(x)cos2x + e^(x)sin2x`

`(dc)/dx = e^(x)(d/dx cos2x) + (d/dx e^(x))cos2x`
`\ \ \ \ \ \ = -2e^(x)sin2x + e^(x)cos2x`

Now integrate the above results:

`int \ (ds)/dx \ dx = int \ 2e^(x)cos2x + e^(x)sin2x \ dx`
`=> s= 2I_c + I_s` ... [A]

`int \ (dc)/dx \ dx = int \ -2e^(x)sin2x + e^(x)cos2x \ dx`
`=> c = -2I_s + I_c` ... [B]

2Eq [A] + Eq [B}:

`2s+c=5I_c`
`:. I_c = 1/5(2s+c)`

From [B] we also get:

`c = -2I_s+1/5(2s+c)`
`:. c = -2I_s+2/5s+1/5c`
`:. I_s = 1/5(s-2c)`

Hence we get the two results:

`int \ e^(x)cos2x \ dx = 1/5(2s+c) + C`
`" " = e^(x)/5(2sin2x+cos2x) + C`

`int \ e^(x)sin2x \ dx = 1/5(s-2c) + C`
`" " = e^(x)/5(sin2x -2cos2x ) + C`