How do i integrate this?

2^(x)*cos3x dx

2 Answers
Feb 19, 2018

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

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:
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