Question

How do you find the integral of `(cosx)(e^x)`?

Answer 1

`int \ cosx \ e^x \ dx = 1/2e^x(cosx + sinx) + C`

Explanation:

Let:

`I = int \ cosx \ e^x \ dx`

We can use integration by parts:

Let `{ (u,=cosx, => (du)/dx=-sinx), ((dv)/dx,=e^x, => v=e^x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (cosx)(e^x) \ dx = (cosx)(e^x) - int \ (e^x)(-sinx) \ dx`
`:. I = e^xcosx + int \ e^x \ sinx \ dx` .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to `I`, having exchanged `cosx` for `sinx`, but if we apply IBP a second time then the progress will become clear:

Let `{ (u,=sinx, => (du)/dx=cosx), ((dv)/dx,=e^x, => v=e^x ) :}`

Then plugging into the IBP formula, gives us:

`int \ (sinx)(e^x) \ dx = (sinx)(e^x) - int \ (e^x)(cosx) \ dx`
`:. int \ e^x \ sinx \ dx = e^xsinx - I`

Inserting this result into [A] we get:

`I = e^xcosx + e^xsinx - I + A`

`:. 2I = e^xcosx + e^xsinx + A`
`:. I = 1/2(e^xcosx + e^xsinx) + C`

Answer 2

`inte^x cosx dx = (e^x(cosx+sinx))/2 +C`

Explanation:

Integrate by parts:

`int e^x cosx dx = int cosx d(e^x) = e^xcosx - int e^x d(cosx) = e^x cosx +int e^xsinxdx`

Now integrate again by parts:

`int e^x sinx dx = int sinx d(e^x) = e^xsinx - int e^x d(sinx) = e^x sinx -int e^xcosxdx`

Substituting this equality in the first one we have:

`int e^x cosx dx = e^xcosx+e^xsinx - int e^xcosxdx`

The integral now appears on both sides of the equation and we can now solve for it:

`2int e^x cosx dx = e^xcosx+e^xsinx +C`

and finally:

`inte^x cosx dx = (e^x(cosx+sinx))/2 +C`

Answer 3

Recursively use integration by parts.
After 2 iterations, the negative of the integral will be on the right.
Subtract form to both sides and divide by 2.

Explanation:

Integration by parts:

`intudv = uv - intvdu`

let `u = cos(x)`, then, `dv = e^xdx`, `du = -sin(x)dx`, and `v = e^x`

Substitute into the formula:

`intcos(x)e^xdx=cos(x)e^x-int-sin(x)e^xdx`

`intcos(x)e^xdx=cos(x)e^x+intsin(x)e^xdx" [1]"`

Integration by parts:

`intudv = uv - intvdu`

let `u = sin(x)`, `dv = e^xdx`, then, `du = cos(x)dx`, and `v = e^x`

Substitute into the formula:

`intsin(x)e^xdx = sin(x)e^x - intcos(x)e^xdx`

Substitute into equation [1]:

`intcos(x)e^xdx=cos(x)e^x+sin(x)e^x - intcos(x)e^xdx`

Add `intcos(x)e^xdx` to both sides:

`2intcos(x)e^xdx=cos(x)e^x+sin(x)e^x`

Divide by 2:

`intcos(x)e^xdx=1/2cos(x)e^x+1/2sin(x)e^x`

Don't forget the constant:

`intcos(x)e^xdx=1/2cos(x)e^x+1/2sin(x)e^x+ C`