Solving Integral of (e^3x)cos(2x) and Cos(sqrt.x)

[ludi_srbin]

Homework Statement

I need to find the integral of (e^3x)cos(2x)

Homework Equations

The Attempt at a Solution

I tried using different trig identities for cos(2x) to get a better equation and then tried to do few versions using the integration by parts but they all kept getting more complicated.




Also I need to find cos(sqrt.x). I am not sure how to do the integration by parts since I have only one piece of equation. Just setting X^1/2=u won't work couse du part has x in it which I can't take out of the integral.

 

[ludi_srbin]

I solved the second one but the first one is still killing me.

 

[d_leet]

Use integration by parts, several times(I think twice will do it) for the first one. I think integration by parts might work for the second one too(try integrating 1*dx and differentiation cos(x^1/2 when you do integration by parts), but I'm not completely sure that will work.

 

[ludi_srbin]

Use integration by parts, several times(I think twice will do it) for the first one. I think integration by parts might work for the second one too(try integrating 1*dx and differentiation cos(x^1/2 when you do integration by parts), but I'm not completely sure that will work.

Got the first one too. Thanks for the help. It would help if we actually learned the derivative of cos(2x), but since we didn't and it is nowhere in the book I just assumed we don't need to use it. Well...I certainly do feel better now.

 

[d_leet]

Got the first one too. Thanks for the help. It would help if we actually learned the derivative of cos(2x), but since we didn't and it is nowhere in the book I just assumed we don't need to use it. Well...I certainly do feel better now.

Glad to help, as for the derivative of cos(2x), if you know the derivative of cos(x) and the chain rule you can work out the derivative of cos(2x).

So we have

y=cos(2x)
if u = 2x then
y=cos(u) and
dy/dx={d[cos(u)]/du}*{du/dx}
y'=-sin(u)*{du/dx}
and remember that u=2x so
y'=-sin(2x)*{d[2x]/dx}
y'=-2sin(2x)

 

[ludi_srbin]

Glad to help, as for the derivative of cos(2x), if you know the derivative of cos(x) and the chain rule you can work out the derivative of cos(2x).

So we have

y=cos(2x)
if u = 2x then
y=cos(u) and
dy/dx={d[cos(u)]/du}*{du/dx}
y'=-sin(u)*{du/dx}
and remember that u=2x so
y'=-sin(2x)*{d[2x]/dx}
y'=-2sin(2x)

Yup. Today in class I realized how stupid was what I wrote here last night.

I guess, for some reason, I assumed that because there is that 2x regular cos formula for derivative doesen't apply.

Thanks for help. I appreciate it.