Question

please integrate the e^(2x) cos3x dx

Answer 1

This uses the product rule, if you know that.

Answer 2

I'm just doing the question.

Answer 3

sorry, it says integrate, hang on

Answer 4

Integrating by parts

Answer 5

you will have to do integration by parts twice

Answer 6

Or use the shortcut method.

Answer 7

please can you show me how to do it?

Answer 8

@tommo_lcfc , is that the tabular method you're referring to?

Answer 9

Tell me what it is and I'll tell you if I'm referring to it or not. I was taught this method at university.

Answer 10

@tommo_lcfc..what kind of method is that?

Answer 11

Basically, if f(x) is a polynomial, we just differentiate this until it becomes constant and integrate the other term.

Answer 12

I understand that the question does not refer to a polynomial but the trig differentiation is repeated. You always get either sin, cos, -sin or -cos (unless you are differentiating something else).

Answer 13

Let's see what you all get.

Answer 14

Let u = e^2x and dv = cos(3x)dx.
Then, du = 2e^2x dx and v = sin(3x)/3

The integral may be rewritten as follows ∫u dv = uv - ∫v du

∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)∫e^(2x) sin(3x) dx T

his requires us to do integration by parts again
p = e^(2x), dp = 2e^(2x) dx, dq = sin(3x) dx, q = -cos(3x)/3

Now, ∫p dq = pq - ∫q dp

∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)[-e^(2x) cos(3x)/3 + (2/3)∫e^(2x) cos(3x) dx]

= e^(2x) sin(3x)/3 + (2/9)e^(2x) cos(3x) - (4/9)∫e^(2x) cos(3x) dx

You have 9/9 of the original integral on the LHS and -4/9 of it on the RHS.

So add 4/9 to both sides (13/9)∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 + (2/9)e^(2x) cos(3x) Finally, multiply both sides by 9/13 and add the constant of integration to get the final answer ∫ e^(2x) cos(3x) dx = (3/13) e^(2x) sin(3x) + (2/13) e^(2x) cos(3x) + C


this is their answer but i was confused in this equation:
∫ e^(2x) cos(3x) dx = e^(2x) sin(3x)/3 - (2/3)[-e^(2x) cos(3x)/3 + (2/3)∫e^(2x) cos(3x) dx]

where did he get the sin(3x)/3

please can you explain it to me?

Answer 15

sin(3x)/3 is what you get when you integrate sin(3x) as you divide the component inside the x term, in this case 3.

Answer 16

Also, I didn't get that answer. I got 3e^(2x) [5/4 sin(3x) - 15/8 cos(3x)] + c using my method.

Answer 17

\[\huge{\int\limits{e^{2x}\cos(3x)dx}}\]

let u=cos(3x) dv=e^2xdx
du=-3sin(3x)dx v=e^2x/2

\[\int\limits{e^{2x}\cos(3x)}=\cos(3x) \frac{e^{2x}}{2}-\int\limits{\frac{e^{2x}}{2}\times (-3\sin(3x))dx}\]\[\int\limits{e^{2x}\cos(3x)dx}=\frac{1}{2}\cos(3x)e^{2x}+\frac{3}{2}\int\limits{e^{2x}\sin(3x)dx}\]
now u do the second integral by parts
\[\int\limits{e^{2x}\sin(3x)dx}\]
let u=sin(3x) dv=e^2xdx
du=3cos(3x)dx v=e^2x/2

so\[\int\limits{e^{2x}\sin(3x)dx}=\sin(3x)\frac{e^{2x}}{2}-\frac{3}{2} \int\limits{e^{2x}\cos(3x)dx}\]
now substitute this in the second equation i wrote\[\int\limits{e^{2x}\cos(3x)dx}=\frac{1}{2}\cos(3x)e^{2x}+\frac{3}{2}[\frac{1}{2}\sin(3x)e^{2x}-\frac{3}{2}\int\limits{e^{2x}\cos(3x)dx)}\]now u have the same integral on both sides .. move one on the other side and solve for the integral

Answer 18

i usually take it as formula
http://answers.yahoo.com/question/index?qid=20080723161756AA9xWm6

Answer 19

thats a good idea :D but also everyone needs to know how to do it in case they forgot the formula :D:D

Answer 20

yeah ... it's the general rule :D
same goes to sin bx :D