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


*

*Your integrand $f$ is symmetric around zero.

*The antiderivative $F$ you chose has value zero at zero.

*In your definite integral, the limits of integration are symmetric around zero.


Given these, you will get
$$
\int_{-a}^af(x)\,dx\stackrel{(1,3)}=2\int_0^af(x)\,dx=2[F(a)-F(0)] \stackrel{(2)}= 2F(a).
$$
A: $$\int \frac{dx}{1-x^2} = $$
$$\frac {1}{2} \int \bigg(\frac{1}{1+x} +  \frac{1}{1-x} \bigg) dx       =$$
$$\frac 12\ln\frac{1+x}{1-x}$$Upon Evaluation at the upper and lower limits, for $$a=\sin (5\pi /{12})$$
Note that $$-a=\sin (-5\pi /{12})$$
Thus $$\frac 12\ln\frac{1+x}{1-x}\bigg|_{-a}^a = $$
$$\frac 12\bigg(\ln\frac{1+a}{1-a} -\ln\frac{1-a}{1+a}\bigg)=$$
$$\frac 12\ln \bigg(\frac{1+a}{1-a}\bigg)^2=$$
$$\ln \bigg(\frac{1+a}{1-a}\bigg)$$
A: $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\left.\frac12\log\frac{1+x}{1-x}\right|_{\sin\left(-\frac{5\pi}{12}\right)}^{\sin\left(\frac{5\pi}{12}\right)}=\frac12\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}-\frac12\log\frac{1+\sin\left(-\frac{5\pi}{12}\right)}{1-\sin\left(-\frac{5\pi}{12}\right)}=$$$${}$$
$$=\frac12\log\left[\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\cdot\frac{1-\sin\left(-\frac{5\pi}{12}\right)}{1+\sin\left(-\frac{5\pi}{12}\right)}\right]\;(**)$$
But $\;\sin(-x)=-\sin x\;$ , so...
$$(**)=\frac12\log\left(\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\right)^2=\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}$$
A: 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.
A: 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.
