TITLE: 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 [5 upvotes]: 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?

REPLY [5 votes]: 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
$$
\int_{-a}^af(x)\,dx\stackrel{(1,3)}=2\int_0^af(x)\,dx=2[F(a)-F(0)] \stackrel{(2)}= 2F(a).
$$