Question

Solve the equation.
$\int \sqrt{3-2x-2x^2}dx$

Answer 1

$\int \sqrt{3-2x-2x^2}dx$

$=\int \sqrt{-(2x^2+2x-3)}dx$

$=\int \sqrt{-2(x^2+x-\frac{3}{2})}dx$

$=\int \sqrt{-2(x^2+x+\frac{1}{4}-\frac{1}{4}-\frac{3}{2})}dx$

$=\int \sqrt{-2(x^2+x+\frac{1}{4}-\frac{1}{4}-\frac{6}{4})}dx$

$=\int \sqrt{-2({(x+\frac{1}{2})}^2-\frac{7}{4})}dx$

$=\int \sqrt{2({\left(\frac{\sqrt{7}}{2}\right)}^2-{(x+\frac{1}{2})}^2)}dx$

$=\sqrt{2}\int \sqrt{({\left(\frac{\sqrt{7}}{2}\right)}^2-{(x+\frac{1}{2})}^2)}dx$

We know that $\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}a{\sin }^{-1}\frac{x}{a}+c$

$=\frac{1}{2}\times \sqrt{2}(x+\frac{1}{2})\sqrt{{\left(\frac{\sqrt{7}}{2}\right)}^2-{(x+\frac{1}{2})}^2}+\frac{1}{2}\left(\frac{\sqrt{7}}{2}\right){\sin }^{-1}\frac{(x+\frac{1}{2})}{\left(\frac{\sqrt{7}}{2}\right)}+c$

$=\frac{\sqrt{2}}{2}\left(\frac{2x+1}{2}\right)\sqrt{\frac{7}{4}+{\left(\frac{2x+1}{2}\right)}^2}+\frac{\sqrt{7}}{4}{\sin }^{-1}\frac{2x+1}{\sqrt{7}}+c$

$=\frac{\sqrt{2}(2x+1)}{8}\sqrt{7+4x^2+1+4x}+\frac{\sqrt{7}}{4}{\sin }^{-1}\frac{2x+1}{\sqrt{7}}+c$

$=\frac{\sqrt{2}(2x+1)}{8}\sqrt{4x^2+4x+8}+\frac{\sqrt{7}}{4}{\sin }^{-1}\frac{2x+1}{\sqrt{7}}+c$