Question

How do I find the integral `int(cos(x)/e^x)dx` ?

Answer 1

`int{cosx}/e^xdx={e^{-x}}/2(sinx-cosx)+C`

Let us look at some details.

Let

`I=int{cos x}/{e^x}dx`

by rewritng `1/e^x` as `e^{-x}`,

`=int e^{-x}cos x dx`

by Integration by Parts

Let `u=e^{-x}` and `dv=cosx dx`
`Rightarrow du=-e^{-x}dx` and `v=sin x`

`=e^{-x}sinx+int e^{-x}sinx dx`

by another Integration by Parts,

Let `u=e^{-x}` and `dv=sinx dx`
`Rightarrow du=-e^{-x}dx` and `v=-cosx`

`=e^{-x}sinx-e^{-x}cosx-I`

So, we have

`I=e^{-x}(sinx-cosx)-I`

by adding `I`,

`2I=e^{-x}(sinx-cosx)`

by dividing by 2,

`I={e^{-x}}/2(sinx-cosx)+C`