Question

How do i integrate this?

2^(x)*cos3x dx

Answer 1

`I=(e^(ln(2)x)(3sin(3x)+ln(2)cos(3x)))/((ln(2))^2+3^2)+C`

Explanation:

We want to solve

`I=int2^xcos(3x)dx=inte^(ln(2)x)cos(3x)dx`

Lets try the more general problem

`I_1=inte^(ax)cos(bx)dx`

Where we seek the solution

`I_1=(e^(ax)(bsin(bx)+acos(bx)))/(a^2+b^2)+C`

The trick is to use integration by parts twice

`intudv=uv-intvdu`

Let `u=e^(ax)` and `dv=cos(bx)dx`

Then `du=ae^(ax)dx` and `v=1/bsin(bx)`

`I_1=1/be^(ax)sin(bx)-a/binte^(ax)sin(bx)dx`

Apply integration by parts to the remaining integral

`I_2=a/binte^(ax)sin(bx)dx`

Let `u=e^(ax)` and `dv=sin(bx)dx`

Then `du=ae^(ax)dx` and `v=-1/bcos(bx)`

`I_2=a/b(-1/be^(ax)cos(bx)+a/binte^(ax)cos(bx)dx)`

`=-a/b^2e^(ax)cos(bx)+a^2/b^2inte^(ax)cos(bx)dx`

`=-a/b^2e^(ax)cos(bx)+a^2/b^2I_1`

Substitute this into the original integral and solve for `I_1`,
it's a bit long, but we take it step by step

`I_1=1/be^(ax)sin(bx)-(-a/b^2e^(ax)cos(bx)+a^2/b^2I_1)`

`I_1=1/be^(ax)sin(bx)+a/b^2e^(ax)cos(bx)-a^2/b^2I_1`

`I_1+a^2/b^2I_1=1/be^(ax)sin(bx)+a/b^2e^(ax)cos(bx)+C`

`(a^2+b^2)/b^2I_1=1/be^(ax)sin(bx)+a/b^2e^(ax)cos(bx)+C`

`I_1=b^2/(a^2+b^2)(1/be^(ax)sin(bx)+a/b^2e^(ax)cos(bx))+C`

`I_1=1/(a^2+b^2)(be^(ax)sin(bx)+ae^(ax)cos(bx))+C`

`I_1=(e^(ax)(bsin(bx)+acos(bx)))/(a^2+b^2)+C`

For your problem `a=ln(2)` and `b=3`

`I=(e^(ln(2)x)(3sin(3x)+ln(2)cos(3x)))/((ln(2))^2+9)+C`

Hopefully there aren't to many mistakes

Answer 2

See the answer below: we have solved using discrete elements instead of a general formulation and we did not simplify the final result, as follows: