Question

Find (and prove) a general formula for integration of a^(bx) cos(cx) dx for any constants a, b, c?

Answer 1

`int a^(bx)cos(cx)dx = a^(bx)(bln(a)cos(cx) + c sin(cx))/( c^2+b^2ln^2(a))`

Explanation:

Let:

`I = int a^(bx)cos(cx)dx`

As:

`a^(bx) = (e^lna)^(bx) = e^(blnax)`

write the integrand as:

`I = int e^(blnax)cos(cx)dx`

Integrate now by parts:

`I = 1/(blna) int cos(cx) d/dx (e^(blnax)) dx`

`I= (e^(blnax)cos(cx))/(blna) -1/(blna) int e^(blnax) d/dx (cos(cx)) dx`

`I = (a^(bx)cos(cx))/(blna) + c/(blna) int e^(blnax) sin(cx) dx`

and integrate by parts again:

`I = (a^(bx)cos(cx))/(blna) + (c e^(blnax) sin(cx))/(blna)^2 - c^2/(blna)^2int e^(blnax) cos(cx) dx`

The same integral now appears in both sides and we can solve for it:

`I = (a^(bx)cos(cx))/(blna) + (c a^(bx) sin(cx))/(blna)^2 - (c^2I)/(blna)^2`

`I (1+c^2/(blna)^2)= (a^(bx)cos(cx))/(blna) + (ca^(bx) sin(cx))/(blna)^2`

`I (c^2+(blna)^2)= a^(bx)(blna)^2(cos(cx)/(blna) + (c sin(cx))/(blna)^2)`

`I = a^(bx)(blnacos(cx) + c sin(cx))/( c^2+(blna)^2)`

Answer 2

`=>int a^(bx)cos(cx)dx= ( (b ln(a) cos(c x) + c sin(c x))a^(b x))/((b ln(a))^2 + c^2)`

Explanation:

`int a^(bx) cos(cx) dx`

First round of integration by parts:

Let `u equiv a^(bx)` and `dv equiv cos(cx) dx`

Then `du = b a^(b x) ln(a) dx " "` and `" "v = 1/c sin(cx)`

`int a^(bx) cos(cx) dx= uv - int v du`

`int a^(bx) cos(cx) dx= 1/ca^(bx)sin(cx) - (bln(a))/(c)int a^(bx)sin(cx)dx`

Second round of integration by parts:

Let `u equiv a^(bx)` and `dv equiv sin(cx)dx`

Then `du = b a^(b x) ln(a) dx " "` and `" "v = -1/c cos(cx)`

`int a^(bx) cos(cx) dx= 1/ca^(bx)sin(cx) - (bln(a))/(c)[uv-int v du]`

`int a^(bx) cos(cx) dx= 1/ca^(bx)sin(cx) - (bln(a))/(c)[-1/ca^(bx)cos(cx)+(bln(a))/(c) int a^(bx)cos(cx)dx]`

`int a^(bx) cos(cx) dx= 1/ca^(bx)sin(cx) +(bln(a))/(c^2)a^(bx)cos(cx)-((bln(a))/(c))^2 int a^(bx)cos(cx)dx`

We can notice that the integral we have just come across is the same as the original integral we wanted to solve for (shown below):

`color(blue)(int a^(bx) cos(cx) dx)= 1/ca^(bx)sin(cx) +(bln(a))/(c^2)a^(bx)cos(cx)-((bln(a))/(c))^2 color(blue)(int a^(bx)cos(cx)dx)`

We can now move this and extract the integral:

`(1+((bln(a))/(c))^2)int a^(bx)cos(cx)dx = 1/ca^(bx)sin(cx) +(bln(a))/(c^2)a^(bx)cos(cx)`

`int a^(bx)cos(cx)dx = {1/ca^(bx)sin(cx) +(bln(a))/(c^2)a^(bx)cos(cx)}/{1+((bln(a))/(c))^2}`

Simplifying to a nicer form:

`=>int a^(bx)cos(cx)dx= ( (b ln(a) cos(c x) + c sin(c x))a^(b x))/((b ln(a))^2 + c^2)`