Question

How do you find the integral of `f(x)=e^(2x)sin3x` using integration by parts?

Answer 1

`1/13e^(2x)(2sin(3x) - 3cos(3x)) + c`

Explanation:

So, with integration by parts we state that:

`intudv = uv - intvdu`

Let's use `e^(2x)` as our u function and `sin3x` as our v' function, leaving us with:

`u = e^(2x)`
`u' = 2e^(2x)`

`v = -1/3cos3x`
`v' = sin3x`

Now let's plug this back into our original formula:

`e^(2x)(-1/3cos3x) - int2e^(2x)(-1/3cos3x)dx`

Let's simplify it a bit more:

`-1/3e^(2x)(cos3x) - (-2/3)inte^(2x)(cos3x)dx`

We have to apply integration by parts yet again, using:

`u = e^(2x)`
`u' = 2e^(2x)`

`v'=cos(3x)`
`v=1/3sin(3x)`

Ultimately giving us:

`-1/3e^(2x)cos(3x)-(-2/3(e^(2x)1/3sin(3x)-int2e^(2x)1/3sin(3x)dx))`

Now, if we isolate `intsin(3x)e^(2x)dx` from this equation we'll end up with:

`intsin(3x)e^(2x)dx = 1/13e^(2x)(2sin(3x)-3cos(3x)) + C`

I'm taking Calc 2 as well, so if anyone has a correction please don't hesitate! Also I'm running late for class so I kinda glossed over the last step, I can explain further if needed!