<s>[INST] 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?
 [/INST] $$\int \frac{dx}{1-x^2} = $$
$$\frac {1}{2} \int \bigg(\frac{1}{1+x} +  \frac{1}{1-x} \bigg) dx       =$$
$$\frac 12\ln\frac{1+x}{1-x}$$Upon Evaluation at the upper and lower limits, for $$a=\sin (5\pi /{12})$$
Note that $$-a=\sin (-5\pi /{12})$$
Thus $$\frac 12\ln\frac{1+x}{1-x}\bigg|_{-a}^a = $$
$$\frac 12\bigg(\ln\frac{1+a}{1-a} -\ln\frac{1-a}{1+a}\bigg)=$$
$$\frac 12\ln \bigg(\frac{1+a}{1-a}\bigg)^2=$$
$$\ln \bigg(\frac{1+a}{1-a}\bigg)$$
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