Question

$\int \frac{1}{\sqrt{1-4x^2}}dx=$

• A. ${\sin }^{-1}2x+c$
• B. $\frac{1}{2}{\sin }^{-1}2x+c$
• C. $({\sin }^{-1}2x{)}^2+c$
• D. $\frac{1}{2}({\sin }^{-1}2x{)}^2+c$

Answer 1

The correct option is B $\frac{1}{2}{\sin }^{-1}2x+c$

Using the integration formula $\int \frac{1}{\sqrt{1-x^2}}dx=si{n}^{-1}x+c$
Now,
$\int \frac{1}{\sqrt{1-4x^2}}dx$

$=\int \frac{1}{\sqrt{1-(2x{)}^2}}dx$

Let $2x=t\Rightarrow dx=\frac{1}{2}dt$

$=\frac{1}{2}\int \frac{1}{\sqrt{1-t^2}}d\left(t\right)=\frac{1}{2}si{n}^{-1}\left(t\right)+c$

$=\frac{1}{2}si{n}^{-1}\left(t\right)+c=\frac{1}{2}si{n}^{-1}\left(2x\right)+c$

$\therefore \int \frac{1}{\sqrt{1-4x^2}}dx=\frac{1}{2}{\sin }^{-1}\left(2x\right)+c$