Welcome to our community

Be a part of something great, join today!

Definite integral involving the natural log function

Bmanmcfly

Member
Mar 10, 2013
42
I figured I would just add this new problem over here, rather than starting a new thread.

Im looking to solve integration leading to arctan or arcsin results.

\(\displaystyle \int_{1}^{e}\frac{3dx}{x(1+\ln(x)^2})\)

Looking at this, it feels like this has an arctan in the result, but I would have to multiply the x with the \(\displaystyle (1+\ln(x)^2)\) and then would figure out a difference of squares to figure out the integral...

that ln(x) is what is screwing me up...

So, I guess I'm asking how working with the ln(x)^2 would be different from working with just an x.

Side question, what is the difference really between \(\displaystyle tan^{-1}, Cot and arctan\) it seems to me that these are just different ways of saying cos/sin. Is there something else significant that I'm ignorant about?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We actually prefer that you begin a new topic for a new question rather than tag a new question onto an existing topic, as per rule #8. I have split the topic so that your new question has its own topic. :D

For this definite integral, I recommend using the substitution:

\(\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx\)

Now, being mindful to change the limits in accordance with the substitution, how may you now rewrite the integral?
 

Bmanmcfly

Member
Mar 10, 2013
42
We actually prefer that you begin a new topic for a new question rather than tag a new question onto an existing topic, as per rule #8. I have split the topic so that your new question has its own topic. :D

For this definite integral, I recommend using the substitution:

\(\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx\)

Now, being mindful to change the limits in accordance with the substitution, how may you now rewrite the integral?

Woops, I was thinking it would keep things more streamline.... Not important.

As for the integral,

Thanks for the pointer, the limits would change to ln(e)- ln(1) or 1-0.
let me know if I got you right...
\(\displaystyle 3\int_{\ln(1)}^{\ln(e)}\frac{(\frac{1}{u}du)}{(1+u^2)}\)

This gives \(\displaystyle 3\tan^{-1}(u)|_{\ln(1)}^{\ln(e)}
\)

Thanks again so much, you don't even know.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You've got the right idea...I would think of the integral as:

\(\displaystyle \int_1^e\frac{3}{x(1+\ln^2(x))}\,dx=3\int_1^e \frac{1}{\ln^2(x)+1}\cdot\frac{1}{x}\,dx\)

Now, when you make the substitution, you may write:

\(\displaystyle 3\int_0^1\frac{1}{u^2+1}\,du=3\left[\tan^{-1}(u) \right]_0^1\)
 

Bmanmcfly

Member
Mar 10, 2013
42
You've got the right idea...I would think of the integral as:

\(\displaystyle \int_1^e\frac{3}{x(1+\ln^2(x))}\,dx=3\int_1^e \frac{1}{\ln^2(x)+1}\cdot\frac{1}{x}\,dx\)

Now, when you make the substitution, you may write:

\(\displaystyle 3\int_0^1\frac{1}{u^2+1}\,du=3\left[\tan^{-1}(u) \right]_0^1\)
Now, the other part of the question;

Is there any significant difference between \(\displaystyle \tan^{-1} , \cot, \and \arctan \)??


It seems that all mean the same thing? What's the reason to use one over the other? Cause to me it seems just preference.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Well, we do have:

\(\displaystyle \arctan(x)\equiv\tan^{-1}(x)\)

However, the cotangent function is the multiplicative inverse of the tangent funtion, not the functional inverse, and this is a common misconception among students.

\(\displaystyle \cot(\theta)\equiv\frac{1}{\tan(x)}\)

whereas if:

\(\displaystyle x=\tan(\theta)\) then \(\displaystyle \theta=\tan^{-1}(x)\)

usually defined where \(\displaystyle -\frac{\pi}{2}<\theta<\frac{\pi}{2}\).

It can be a confusing notation, where for variables, we take:

\(\displaystyle x^{-1}=\frac{1}{x}\)

but for functions:

\(\displaystyle f^{-1}(x)\ne\frac{1}{f(x)}\)

and by definition:

\(\displaystyle f^{-1}\left(f(x) \right)=x\)