Question
Evaluate the following integrals:$\int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$
✓
Answer
$7 - 6x - x^2$ can be written as $7 - (x^2 + 6x + 9 - 9).$
Therefore,$7-(\text{x}^2+6\text{x}+9-9)$
$=16-(\text{x}^2+6\text{x}+9)$ $=16-(\text{x}+3)^2$ $=(4)^2-(\text{x}+3)^2$
$\therefore\ \int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$ $=\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}$
Let $x + 3 = t$
$\Rightarrow\text{dx}=\text{dt}$
$\Rightarrow\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{(4)^2-(\text{t})^2}}\text{ dt}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{4}\Big)+\text{C}$
Therefore,$7-(\text{x}^2+6\text{x}+9-9)$
$=16-(\text{x}^2+6\text{x}+9)$ $=16-(\text{x}+3)^2$ $=(4)^2-(\text{x}+3)^2$
$\therefore\ \int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$ $=\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}$
Let $x + 3 = t$
$\Rightarrow\text{dx}=\text{dt}$
$\Rightarrow\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{(4)^2-(\text{t})^2}}\text{ dt}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{4}\Big)+\text{C}$
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