Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 62

Evaluate the integral $\displaystyle \int \frac{dx}{\sqrt{1-4t^2}}$
If we let $u = 2x$, then $du = 2dx$
Thus,


$\begin{equation} \begin{aligned} \int \frac{dx}{\sqrt{1-(2x)^2}} &= \frac{1}{2} \int \frac{du}{\sqrt{1-u^2}}\\ \\ &= \frac{1}{2} \sin^{-1}(u) + c\\ \\ &= \frac{1}{2} \sin^{-1}(2x) + c \end{aligned} \end{equation}$