1b. ∫ 1 / √(1 - 4x²) dx

Explanation:

Step1: Use substitution

Let $u = 2x$, then $du=2dx$ and $dx=\frac{1}{2}du$.

Step2: Rewrite the integral

The integral $\int\frac{1}{\sqrt{1 - 4x^{2}}}dx$ becomes $\int\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{1}{2}du=\frac{1}{2}\int\frac{1}{\sqrt{1 - u^{2}}}du$.

Step3: Recall the antiderivative formula

We know that $\int\frac{1}{\sqrt{1 - t^{2}}}dt=\arcsin(t)+C$. So, $\frac{1}{2}\int\frac{1}{\sqrt{1 - u^{2}}}du=\frac{1}{2}\arcsin(u)+C$.

Step4: Substitute back

Substitute $u = 2x$ back, we get $\frac{1}{2}\arcsin(2x)+C$.

Answer:

$\frac{1}{2}\arcsin(2x)+C$