# Why $\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$?

Why does an integral $$\int \frac{dx}{1-x^2}$$ with the limitless (undefined) interval equal to $$\frac 12\ln\frac{1+x}{1-x},$$ yet an integral $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}$$ with an interval from $\sin\frac{-5π}{12}$ to $\sin\frac{5π}{12}$ has $$\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$$ without one half attached to ln?

• use MathJax for better formatting Commented Mar 29, 2018 at 11:13
• What? Not only it's very hard to understand without proper formatting, but also the wording is pretty confusing... Commented Mar 29, 2018 at 11:15

$$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\left.\frac12\log\frac{1+x}{1-x}\right|_{\sin\left(-\frac{5\pi}{12}\right)}^{\sin\left(\frac{5\pi}{12}\right)}=\frac12\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}-\frac12\log\frac{1+\sin\left(-\frac{5\pi}{12}\right)}{1-\sin\left(-\frac{5\pi}{12}\right)}=$$$${}$$

$$=\frac12\log\left[\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\cdot\frac{1-\sin\left(-\frac{5\pi}{12}\right)}{1+\sin\left(-\frac{5\pi}{12}\right)}\right]\;(**)$$

But $\;\sin(-x)=-\sin x\;$ , so...

$$(**)=\frac12\log\left(\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\right)^2=\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}$$

I don't know but the explanation is really trivial. The second one is a definite integral symmetric around $x=0$ or rather in general you have for some positive $r < 1$,

$$\int_{-r}^{r} \frac{dx}{1-x^2} = \frac 12\ln\frac{1+x}{1-x} \Bigg|^{r}_{-r} = \ln\frac{1+r}{1-r}$$

Hope it helps.

• Why does your answer start with "I don't know but", when it seems that you do know? Commented Mar 29, 2018 at 22:35
• This answer removes the fluff and gets right to the heart of the matter. Other than the awkward initial phrase, I think it is exemplary. So much so that I commented to that effect! Commented Mar 30, 2018 at 0:44

You gained a factor of two because three conditions were satisfied:

1. Your integrand $f$ is symmetric around zero.
2. The antiderivative $F$ you chose has value zero at zero.
3. In your definite integral, the limits of integration are symmetric around zero.

Given these, you will get $$\int_{-a}^af(x)\,dx\stackrel{(1,3)}=2\int_0^af(x)\,dx=2[F(a)-F(0)] \stackrel{(2)}= 2F(a).$$

• 2... which is always possible since the integration constant is arbitrary. Commented Mar 29, 2018 at 18:30
• @Miguel Agreed, but if OP had chosen $F$ with $F(0)\ne 0$ then the definite integral would not equal $2F(a)$. Commented Mar 29, 2018 at 18:44
• Sure, I was only trying to highlight some general procedure beyond the particular example. Commented Mar 29, 2018 at 18:46

$$\int \frac{dx}{1-x^2} =$$

$$\frac {1}{2} \int \bigg(\frac{1}{1+x} + \frac{1}{1-x} \bigg) dx =$$

$$\frac 12\ln\frac{1+x}{1-x}$$Upon Evaluation at the upper and lower limits, for $$a=\sin (5\pi /{12})$$

Note that $$-a=\sin (-5\pi /{12})$$

Thus $$\frac 12\ln\frac{1+x}{1-x}\bigg|_{-a}^a =$$

$$\frac 12\bigg(\ln\frac{1+a}{1-a} -\ln\frac{1-a}{1+a}\bigg)=$$

$$\frac 12\ln \bigg(\frac{1+a}{1-a}\bigg)^2=$$

$$\ln \bigg(\frac{1+a}{1-a}\bigg)$$

• Of course it doesn't affect your answer, but the limits in the problem have $a = \sin(5\pi/12)$, not $a = \sin(\pi/12)$. Commented Mar 29, 2018 at 22:47
• Thanks for the comment. I have edited my answer thanks to your comment. Commented Mar 29, 2018 at 22:54

Since there are already good answers explaining the computation of the definite integral, let me point out what I consider a misunderstanding of the indefinite integral, thus answering the OP question why.

The primitive or indefinite integral is not a function, but a representation of a whole set of functions, all of them when differentiated produce the original function under the integral sign. All these functions differ by a constant, and that is why the computation of the indefinite integral is usually expressed by choosing any of these primitives plus a constant $C$, i.e.: $$\int \frac{dx}{1-x^2}=\frac 12\ln\frac{1+x}{1-x}+C$$ but you could also write: $$\int \frac{dx}{1-x^2}=\frac 12\ln \left(2\frac{1+x}{1-x}\right)+C$$ since these two primitives differ by a constant.

Now you can compute a definite integral by Barrow's rule: $$\int_a^bf(x) dx=F(b)-F(a)$$ where $F'(x)=f(x)$, i.e. $F$ is any primitive or, if you want, the primitive with any value of the constant $C$.

Since you are free to choose the value of $C$, the "aspect" of the primitive can be deceiving with respect to the value of the definite integral, unless you carry out all the detailed computation.

• Sooooo frustrating unexplained downvoting :( Commented Mar 29, 2018 at 18:13
• I didn't downvote, but it seems that this answer only addresses a potential confusion regarding definite integration in general, not anything to do with the poster's specific confusion. (Also, I might argue that one could say instead, not that the indefinite integral is not a function, but rather that there is no such thing as the indefinite integral; rather there are many, each a function, and any two of them differing by a constant.) Commented Mar 29, 2018 at 22:48