Question

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

Answer 1

Consider the given integral.

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

Let $t=4-x^2$

$\frac{dt}{dx}=0-2x$

$-\frac{dt}{2}=xdx$

Therefore,

$I=-\frac{1}{2}\int \frac{1}{\sqrt{t}}dt$

$I=-\frac{1}{2}\left(2\sqrt{t}\right)+C$

$I=-\sqrt{t}+C$

On putting the value of $t$, we get

$I=-\sqrt{4-x^2}+C$

Hence, this is the answer.