Integrate e^2x cos (e^2x) (1 Viewer)

[ngoetz]

Hi, I was wondering if somebody could please help me integrate Integrate e^2x cos (e^2x).
Its on page 532 of Cambridge Y11 3U

Also, Integrate (sin(e^(-2x))/(e^2x)

http://www.wolframalpha.com/input/?i=(sin(e^(-2x))/(e^2x)

Thank you

 

[Drongoski]

Hi, I was wondering if somebody could please help me integrate Integrate e^2x cos (e^2x).
Its on page 532 of Cambridge Y11 3U

Also, Integrate (sin(e^(-2x))/(e^2x)

http://www.wolframalpha.com/input/?i=(sin(e^(-2x))/(e^2x)

Thank you

Sorry still don't know where/how to access LaTeX


1) int [e^2x cose^2x dx = 0.5 int cos e^2x d e^2x = 0.5 sin e^2x + C


2) int ( sin e^-2x)/e^-2x = int sin e^-2x e^-2x dx = -0.5 int sin e^-2x de^-2x

= +0.5 cos e^-2x + C

 

[deterministic]

Sorry still don't know where/how to access LaTeX


1) int [e^2x cose^2x dx = 0.5 int cos e^2x d e^2x = 0.5 sin e^2x + C


2) int ( sin e^-2x)/e^-2x = int sin e^-2x e^-2x dx = -0.5 int sin e^-2x de^-2x

= +0.5 cos e^-2x + C

+1 very nice

OP: if you're not sure what drong0ski did, try using substitutions instead, with u=e^2x for the first one, u=e^(-2x) for the second one

 

[iSplicer]

+1 very nice

OP: if you're not sure what drong0ski did, try using substitutions instead, with u=e^2x for the first one, u=e^(-2x) for the second one

He did exactly the same thing, but in a more elegant way --> instead of taking u = e^blah, he modified the differential change instead, making it d(e^blah)

 

[ngoetz]

Sorry still don't know where/how to access LaTeX


1) int [e^2x cose^2x dx = 0.5 int cos e^2x d e^2x = 0.5 sin e^2x + C


2) int ( sin e^-2x)/e^-2x = int sin e^-2x e^-2x dx = -0.5 int sin e^-2x de^-2x

= +0.5 cos e^-2x + C

Thanks a lot, but I'm not quite sure how you got to the bit in bold? Is that trig substitution?

 

[Drongoski]

In this case I'm doing what is normally done with substitution without the bother of the process. For this you have to learn that d f(x) = f'(x)dx; looking at this in reverse f'(x)dx = d f(x) a more general form of the DIFFERENTIAL - you are almost learning only the most basic form dx, du, dt etc.

When I see xdx I see it as potential .5 dx² , [1/4]d(2x² - 7) , -[1/14] d(7x^2 + 1) . Which one we use depends on the situation at hand.

Thus in our case e^2xdx = [1/2]d e^2x

Similarly e^x sin e^x dx = sin e^x de^x = dsin e^x = d [sin e^x - 5] if this last were at all useful.

All this would look much nicer using LaTeX.

 

[Aquawhite]

Sorry still don't know where/how to access LaTeX


1) int [e^2x cose^2x dx = 0.5 int cos e^2x d e^2x = 0.5 sin e^2x + C


2) int ( sin e^-2x)/e^-2x = int sin e^-2x e^-2x dx = -0.5 int sin e^-2x de^-2x

= +0.5 cos e^-2x + C

http://www.codecogs.com/latex/eqneditor.php

Use it.

 

[Drongoski]

It's embarassing - but how? Childlike - step-by-step what do I have to do. While trying to post what steps must I follow to access LaTeX. Under precious arrangement I've been using it extensively. I'm not like your generation - very well-versed in the ways of the web.

 

[Pwnage101]

It's embarassing - but how? Childlike - step-by-step what do I have to do. While trying to post what steps must I follow to access LaTeX. Under precious arrangement I've been using it extensively. I'm not like your generation - very well-versed in the ways of the web.

Not sure if there's an easier way, but you could:

1). Open the provided link http://www.codecogs.com/latex/eqneditor.php (perhaps bookmark this)

2). Use this equation editor

3). When finished, copy the text into the reply box, with "[_tex]" before and "[_/tex]" [with no underscore ("_")] after the code copied from the website.

e.g. "[_tex] \frac{a}{b} [_/tex]" without the underscores ("_") would produce:

 

[Drongoski]

Ah! Thanks. I'll look at this later pwnage.

 

[Aquawhite]

Whichever method is easiest I guess. I was actually using the latex editor but using the [_tex] input instead of copying the img.

 

[Drongoski]

Now that I know I can input in TeX directly it's easier for me now. I just have to read up on TeX. Thanks for your help Aquawhite.

 

[Trebla]

There are two approaches one can use. Reverse chain rule or formal substitution of u = e^2x.

Reverse chain rule uses the fact that:



The integral is



Which can be expressed as



By observation you should be able to 'recognise' the chain rule derivative in the integrand where f ' is the cos(...) function and g' is the 2e^2x. So by inspection it becomes



A similar approach can be used for the other integral...

 

[ngoetz]

There are two approaches one can use. Reverse chain rule or formal substitution of u = e^2x.

Reverse chain rule uses the fact that:



The integral is



Which can be expressed as



By observation you should be able to 'recognise' the chain rule derivative in the integrand where f ' is the cos(...) function and g' is the 2e^2x. So by inspection it becomes



A similar approach can be used for the other integral...

Thanks that cleared it up a lot

 

[hscishard]

Sorry still don't know where/how to access LaTeX


1) int [e^2x cose^2x dx = 0.5 int cos e^2x d e^2x = 0.5 sin e^2x + C


2) int ( sin e^-2x)/e^-2x = int sin e^-2x e^-2x dx = -0.5 int sin e^-2x de^-2x

= +0.5 cos e^-2x + C

that is elite man...

 

[Drongoski]

that is elite man...

Very kind of you; thank you.