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} $

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now