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

Question

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?

Answer 1

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

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

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

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

Answer 2

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.

Answer 3

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

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

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

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

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

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

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

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

Answer 4

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.

Answer 5

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