MHB Natural log and trig function question

[DeusAbscondus]

when I differentiate:
$$ln4x+sin(x)$$
I get:
$$\frac{1}{x}+cos(x)$$
and Wolfgram agrees

But then when i test this by calculating indefinite integral, I get:
$$ln(x)+cos(x)$$
Which leaves me with three questions:
1. what happened to the 4?
2. why isn't it integrating back to (at least) ln(x)+sin(x)?
3. why doesn't wolfram add $+C$ to the end of an indefinite integral,
seemingly defying a principle over which I've had my knuckles wrapped severely (and liked it) ?

Thanks,

Deus Abscondus or:
"God has absconded from the scene ... again!"

 

[MarkFL]

Hint: $\displaystyle \ln|4x|=\ln|x|+2\ln(2)$

The constant $\displaystyle 2\ln(2)$ gets lumped in with the constant of integration since it is just a constant itself.

I believe Wolfram states + constant on the end of indefinite integrals.

 

[MarkFL]

...
You have differentiated \(\ln(4x)\) incorrectly.

Perhaps he applied the chain rule and simplified...he does have the correct result. :)

 

[Sudharaka]

when I differentiate:
$$ln4x+sin(x)$$
I get:
$$\frac{1}{x}+cos(x)$$
and Wolfgram agrees

But then when i test this by calculating indefinite integral, I get:
$$ln(x)+cos(x)$$
Which leaves me with three questions:
1. what happened to the 4?
2. why isn't it integrating back to (at least) ln(x)+sin(x)?
3. why doesn't wolfram add $+C$ to the end of an indefinite integral,
seemingly defying a principle over which I've had my knuckles wrapped severely (and liked it) ?

Thanks,

Deus Abscondus or:
"God has absconded from the scene ... again!"

Hi DeusAbscondus, :)

Sorry about the obvious error in my last post, I have deleted to avoid any confusion. >>Here<< is what Wolfram gives me. Note that there is the constant term and also the answer is \(\ln(x)+\sin(x)\). :)

Kind Regards,
Sudharaka.

 

[DeusAbscondus]

Thanks Sudharaka,

Well that would be:
$$\frac{d}{dx}ln(4x)+sin(x)=4*\frac{1}{x}+cos(x)$$ I suppose

But then, when you integrate this: you get $$4ln(x)-sin(x)+C$$
don't you? Which is patently not the original $$f(x)=ln4x+sin(x)$$

 

[Sudharaka]

Thanks Sudharaka,

Well that would be:
$$\frac{d}{dx}ln(4x)+sin(x)=4*\frac{1}{x}+cos(x)$$ I suppose

But then, when you integrate this: you get $$4ln(x)-sin(x)+C$$
don't you? Which is patently not the original $$f(x)=ln4x+sin(x)$$

I am sorry about this but the answer you have obtained after differentiating \(\ln(4x)+\sin(x)\) in post #1 is correct and I was wrong in my post #2 and hence deleted it.

\[\frac{d}{dx}ln(4x)+sin(x)=\frac{1}{x}+cos(x)\]

as you have correctly stated in your original post. Refer post #4 and follow the link to see what Wolfram gives after integrating. Then look at Mark's post #2 to clarify more about what happens to the \(4\) as you have asked in your original post. If you have any more questions please don't hesitate to ask. :)

 

[DeusAbscondus]

Hint: $\displaystyle \ln|4x|=\ln|x|+2\ln(2)$The constant $\displaystyle 2\ln(2)$ gets lumped in with the constant of integration since it is just a constant itself.I believe Wolfram states + constant on the end of indefinite integrals.

Forgive my obtuseness: but I still can't see why the integration seems to lose the constant 4.Though I understand and appreciate the nice:: $$\displaystyle \ln|4x|=\ln|x|+2\ln(2)$$Btw, remind me again: what does the command: \displaystyle do?

 

[Sudharaka]

Forgive my obtuseness: but I still can't see why the integration seems to lose the constant 4.Though I understand and appreciate the nice:: $$\displaystyle \ln|4x|=\ln|x|+2\ln(2)$$

\[\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(x)+ \sin(x)+A\]

where \(A\) is the arbitrary constant.

As Mark had pointed out you can write,

\[\ln(4x)=\ln(x)+2\ln(2)\Rightarrow \ln(x)=\ln(4x)-2\ln(2)\]

Substitute this in the above equation and we get,

\[\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(4x)+ \sin(x)+(A-2\ln(2))\]

Now \(A-2\ln(2)\) is also an arbitrary constant and we label it as \(C\).

\[\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(4x)+ \sin(x)+C\]

So you see it is irrelevant whether you write \(\ln(x)+\sin(x)+A\) or \(\ln(4x)+\sin(x)+C\) because both answers are the same and one can be obtained from the other.

 

[Sudharaka]

Btw, remind me again: what does the command: \displaystyle do?

It is a size command. See http://www.mathhelpboards.com/f26/latex-tip-displaystyle-283/#post1711 thread for some uses of \displaystyle.

 

[DeusAbscondus]

\[\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(x)+ \sin(x)+A\]

where \(A\) is the arbitrary constant.

As Mark had pointed out you can write,

\[\ln(4x)=\ln(x)+2\ln(2)\Rightarrow \ln(x)=\ln(4x)-2\ln(2)\]

Substitute this in the above equation and we get,

\[\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(4x)+ \sin(x)+(A-2\ln(2))\]

Now \(A-2\ln(2)\) is also an arbitrary constant and we label it as \(C\).

\[\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(4x)+ \sin(x)+C\]

So you see it is irrelevant whether you write \(\ln(x)+\sin(x)+A\) or \(\ln(4x)+\sin(x)+C\) because both answers are the same and one can be obtained from the other.

I love it!
Thanks Sudharaka; now I can relax, take a break, go eat, and digest properly having cleared that up.
Deus Abs

 

[Sudharaka]

I love it!
Thanks Sudharaka; now I can relax, take a break, go eat, and digest properly having cleared that up.
Deus Abs

It's a pleasure to help you. :)

 

[MarkFL]

I have seen quite a few students run into this same issue with logs regarding anti-derivatives. This log property comes in handy too when solving certain differential equations.