$\int_{-1}^{1} \frac{1}{\sqrt{-3u²+4}}du$

Question

How would one go about solving

$$\int_{-1}^{1} \frac{1}{\sqrt{-3u^2+4}}du$$

I was told I need to make a substitution with $u=\frac{2}{\sqrt{3}}\sin(t)$ but I don't see where this substitution comes, it would have never occurred to me. Any ideas?

Answer 1

The idea is to write the expression under the root as a square of a trigonometric function using the identity $$ \sin^2 t + \cos^2 t = 1. $$

In your case, if $u = 2\sin (t) /\sqrt{3}$ then $u^2 = 4\sin^2(t)/3$ and $$ 4 - 3u^2 = 4 - 4 \sin^2 t = 4 \left(1 - \sin^2 t\right) = 4 \cos^2 t, $$ and $du = 2\cos t dt/\sqrt{3}$, so you end up with $$ \int \frac{du}{\sqrt{4-3u^2}} = \int \frac{2\cos t dt/\sqrt{3}}{\sqrt{4\cos^2 t}} $$ Can you now finish the problem?