Question

How do you integrate `int e^(3x)cos(2x)` by integration by parts method?

Answer 1

I got: `9/13[e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)]+c`
BUT check my maths.

Explanation:

Ok let us try:
`inte^(3x)cos(2x)dx=e^(3x)/3cos(2x)-int(e^(3x)/3)(-2sin(2x))dx=e^(3x)/3cos(2x)+2/3int(e^(3x))sin(2x)dx=`
again:
`=e^(3x)/3cos(2x)+2/3[e^(3x)/3sin(2x)-int(e^(3x)/3)(2cos(2x))dx]`

So basically we get:

`inte^(3x)cos(2x)dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)-4/9inte^(3x)(2cos(2x))dx`

Now a trick....
Let us take `-4/9inte^(3x)(2cos(2x))dx` to the left of the `=` sign:
`inte^(3x)cos(2x)dx+4/9inte^(3x)(2cos(2x))dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)`
add the two integral on the left:
`13/9inte^(3x)(2cos(2x))dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)`
and:
`inte^(3x)(2cos(2x))dx=9/13[e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)]+c`