# natural log and trig function question

#### DeusAbscondus

##### Active member
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!"

Last edited:

#### MarkFL

Staff member
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

Staff member
...
You have differentiated $$\ln(4x)$$ incorrectly.
Perhaps he applied the chain rule and simplified...he does have the correct result.

#### Sudharaka

##### Well-known member
MHB Math Helper
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

##### Active member
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

##### Well-known member
MHB Math Helper
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

##### Active member
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
Btw, remind me again: what does the command: \displaystyle do?
It is a size command. See >>this<< thread for some uses of \displaystyle.

#### DeusAbscondus

##### Active member
$\int\left(\frac{1}{x}+cos(x)\right)\,dx=\ln(x)+ \sin(x)+A$

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

##### Well-known member
MHB Math Helper
I love it!
Thanks Sudharaka; now I can relax, take a break, go eat, and digest properly having cleared that up.
Deus Abs