# Find integral of (tant)/(1+t^2)

1. Dec 1, 2011

### skyturnred

1. The problem statement, all variables and given/known data

I am trying to do the following:

$\int$$^{x^{2}}_{0}$$\frac{tant}{1+t^{2}}$ dt

2. Relevant equations

3. The attempt at a solution

I have only learned substitution rule so far.. I am not sure if this requires trig substitution, or if we are supposed to even know it! We were definitely not taught what trig substitution specifically was, but I am not sure if we were supposed to just "pick it up."

Anyways.. Can someone give me a hint please?

Thanks so much, physics forums have been invaluable recently as I near finals!

2. Dec 1, 2011

### gb7nash

Just out of curiousity, is that supposed to be tant or tan-1t?

3. Dec 1, 2011

### skyturnred

It would make sense for it to be arctant, but after double checking, it's still t. I will write the question from the beginning just in case I made a mistake.

I am supposed to solve the following:

lim x→0 x$^{-8}$ * $\int$$^{x^{2}}_{0}$ $\frac{t}{1+t^{2}}$-$\frac{tant}{1+t^{2}}$ dt

So I am just isolating the integral part first, trying to solve it. I split the integral up into two parts, the integral of the first fraction minus the integral of the second fraction. The first fraction, by substitution, I got it to be $\frac{1}{2}$(ln|1+x$^{4}$|)

The second integral, I can't solve.

4. Dec 2, 2011

### Curious3141

Don't split the integrals. The way to do this is to use FTC/Leibniz Rule, L' Hopital's and the Taylor series for tangent.

Right, the answer I get is (-1/12).

Outline of method:

Treat the problem as the limit of a quotient of two functions, the numerator being that definite integral and the denominator being x^8.

Observe that this is of the indeterminate form 0/0, allowing you to use L' Hopital's Rule.

Use Fundamental Theorem of Calculus (FTC) and Chain Rule [or Leibniz Rule] to differentiate the numerator wrt x. The derivative of the denominator is elementary.

Use the Taylor series of tan(x^2) to express the numerator as a power series in terms of x. You may ignore higher order terms after a certain exponent (this should be obvious).

Group terms, cancel, simplify and take the limit, et voila!

EDIT: Waitaminute. If you've "just" learned substitution, how are you supposed to know enough to do this problem?

Further edit: You can avoid the use of Taylor series by applying L' Hopital's recursively until the denominator and numerator both become nonzero at the limit. You will find that your work is much simplified if you start with a substitution of u = x^2 right at the start and evaluate the limit as u -> 0. This makes the work easier, but is not strictly necessary.

Last edited: Dec 2, 2011
5. Dec 2, 2011

### skyturnred

Thank you very much! That helped me solve it!

They've been doing that to us all year in math, mechanics, physics, etc. I think that they just have too many first year engineers (950+) and are just trying to cut down the numbers as much as possible for next year. Survival of the fittest I guess, which is why you guys are helping me more than you think you are!

6. Dec 2, 2011

### Curious3141

Hi, glad you solved it! :) Please take a look at my further edit - you can avoid Taylor series that way. Just a bit longer.

7. Dec 2, 2011

### lurflurf

recognize
$$\lim_{x \rightarrow 0} \frac{1}{x^8} \int_0^{x^2} \frac{t-\tan(t)}{1+t^2} dt=\lim_{x \rightarrow 0} \frac{d}{d(x^8)} \int_0^{x^2} \frac{t-\tan(t)}{1+t^2} dt=\lim_{x \rightarrow 0} \frac{x^2-\tan(x^2)}{4x^6 (1+x^4)}$$
which can be deduced from
$$\lim_{x \rightarrow 0} \frac{y-\tan(y)}{y^3}=- \frac{1}{3}$$

Last edited: Dec 2, 2011
8. Dec 2, 2011

### skyturnred

Are you sure? The reason why I ask this is because I was able to ask a few other engineers and they all seem to be getting -1/12. I didn't have time to go over their methods though.

9. Dec 2, 2011

### Curious3141

Denominator should be $4x^6(1+x^4)$

10. Dec 2, 2011

### Curious3141

The answer is (-1/12). Lurflurf is simply stating the limit in that last step as (-1/3), whereas we are calculating it (using either Taylor series or L' Hopital's Rule). There's still a factor of (1/4) to be multiplied to get the original limit, which means lurflurf's answer is still (-1/12) [i.e. 1/4*(-1/3)].

11. Dec 2, 2011

### lurflurf

thanks I fixed 1+x^2->1+x^4
it does not effect the limit though
The second limit is not equal to the first, but is related
$$\lim_{x \rightarrow 0} \frac{1}{x^8} \int_0^{x^2} \frac{t-\tan(t)}{1+t^2} dt==\frac{1}{4} \lim_{x \rightarrow 0} \frac{y-\tan(y)}{y^3}=- \frac{1}{4} \cdot \frac{1}{3}=- \frac{1}{12}$$

$$\lim_{x \rightarrow 0} \frac{y-\tan(y)}{y^3}=- \frac{1}{3}$$
can be found by recognizing it as a derivative of t-tan(t) or L'Hôpital's rule or series expansion or some trig identities

Last edited: Dec 2, 2011