Question

$\int \sqrt{3-2x-x^2}dx=$
(where $C$ is integration constant)

• A. ${\sin }^{-1}\left(\frac{x+1}{2}\right)+C$
• B. $\log |x+1+\sqrt{3-2x-x^2}|+C$
• C. $\frac{1}{2}(x+1)\sqrt{3-2x-x^2}+2{\sin }^{-1}\left(\frac{x+1}{2}\right)+C$
• D. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x+1}{\sqrt{2}}\right)+C$

Answer 1

The correct option is C $\frac{1}{2}(x+1)\sqrt{3-2x-x^2}+2{\sin }^{-1}\left(\frac{x+1}{2}\right)+C$

$I=\int \sqrt{3-2x-x^2}dx$
$=\int \sqrt{4-(x+1{)}^2}dx$
Put $x+1=y\Rightarrow dx=dy$, so
$I=\int \sqrt{4-y^2}dy$
We know that,
$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
Therefore,
$I=\frac{1}{2}\cdot y\sqrt{4-y^2}+\frac{4}{2}{\sin }^{-1}\frac{y}{2}+C=\frac{1}{2}(x+1)\sqrt{3-2x-x^2}+2{\sin }^{-1}\left(\frac{x+1}{2}\right)+C$