Question

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

Answer 1

`int e^(3x)cosx dx = e^{3x}/10(3cos(x)+sin(x))`

Explanation:

This kind of integral can be straightforward solved using Moivre's identity

`e^{i x} = cos x + i sin x` so

`int e^(3x)cosx dx = "Real"(int e^(3x)(cosx +i sin x)dx)` or
`int e^(3x)cosx dx = "Real"(int e^{3x+ix}dx) = "Real"(e^{3x+ix}/(3+i))`

then

`int e^(3x)cosx dx ="Real"((3-i)e^{3x}(cosx +i sin x)/((3-i)(3+i)))`

Finally

`int e^(3x)cosx dx = e^{3x}/10(3cos(x)+sin(x))`

Answer 2

We can IBP this both ways

First approach

`I = int e^(3x)cosx \ dx`

`= int d/dx( 1/3e^(3x)) cosx \ dx`

which by IBP

`= 1/3e^(3x) cosx - int 1/3e^(3x) d/dx( cosx )\ dx`

`= 1/3e^(3x) cosx + int 1/3e^(3x) sin x \ dx`

preparing for second IBP

`= 1/3e^(3x) cosx + int d/dx( 1/9e^(3x)) sin x \ dx`

by IBP
`= 1/3e^(3x) cosx + 1/9e^(3x) sin x - int 1/9e^(3x) d/dx (sin x) \ dx`

`= 1/3e^(3x) cosx + 1/9e^(3x) sin x - int 1/9e^(3x) cos \ dx`

`implies I = 1/3e^(3x) cosx + 1/9e^(3x) sin x -I/9 + C`

`I = 9/10 (1/3e^(3x) cosx + 1/9e^(3x) sin x ) + C`

`I = e^(3x)/10( 3 cosx + sin x) + C`

Second Approach

`I = int e^(3x)cosx \ dx`

`= int e^(3x) d/dx(sinx) \ dx`

`= e^(3x) sinx - int d/dx(e^(3x)) sin x \ dx + C`

`= e^(3x) sinx - int 3e^(3x) sin x \ dx + C`

Preparing for second IBP

`= e^(3x) sinx - int 3e^(3x) d/dx(- cos x) \ dx + C`

`= e^(3x) sinx + 3e^(3x) cosx - int d/dx (3e^(3x)) cos x \ dx + C`

`= e^(3x) sinx + 3e^(3x) cosx - int 9e^(3x) cos x \ dx + C`

`= e^(3x) sinx + 3e^(3x) cosx - 9I + C`

`implies 10 I = e^(3x) sinx + 3e^(3x) cosx + C`

`I = e^(3x)/10 ( sinx + 3 cosx ) + C`

Same result, second way maybe a little bit snappier.