Definite integral involving the natural log function

[bmanmcfly]

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]

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]

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]

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]

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]

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:

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