Question
Evaluate the following integrals:
$\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
$\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
✓
Answer
$\text{I}=\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{7}{4}-\Big(\frac{1}{4}+\text{x}+\text{x}^2\Big)}\text{dx}$ $\Big[\text{Adding and subtracting }\frac{1}{4}\Big]$
$=\sqrt2\int\sqrt{\Big(\frac{\sqrt7}{2}\Big)^2-\Big(\text{x}+\frac{1}{2}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\text{x}+\frac{1}{2}}{2}\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}+\frac{7}{8}\sin^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt7}{2}}\bigg)+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{2\text{x}+1}{4}\sqrt{3-2\text{x}-2\text{x}^2}+\frac{7\sqrt2}{8}\sin^{-1}\Big(\frac{2\text{x}+1}{\sqrt7}\Big)+\text{C}$
$=\sqrt2\int\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{7}{4}-\Big(\frac{1}{4}+\text{x}+\text{x}^2\Big)}\text{dx}$ $\Big[\text{Adding and subtracting }\frac{1}{4}\Big]$
$=\sqrt2\int\sqrt{\Big(\frac{\sqrt7}{2}\Big)^2-\Big(\text{x}+\frac{1}{2}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\text{x}+\frac{1}{2}}{2}\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}+\frac{7}{8}\sin^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt7}{2}}\bigg)+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{2\text{x}+1}{4}\sqrt{3-2\text{x}-2\text{x}^2}+\frac{7\sqrt2}{8}\sin^{-1}\Big(\frac{2\text{x}+1}{\sqrt7}\Big)+\text{C}$
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