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Evaluate : $ \int \large\frac{dx}{\sqrt{8+2x-x^2}} $

1 Answer

Toolbox:
  • $\int\large\frac{dx}{\sqrt{a^2-x^2}}$$=\sin^{-1}\big(\large\frac{x}{a}\big)$$+c.$
Step 1:
$I=\int\large\frac{dx}{\sqrt{8+2x-x^2}}$
Consider $8+2x-x^2=-(x^2-2x-8)$
$\qquad\qquad\qquad\qquad=[(x-1)^2-1-8]$
$\qquad\qquad\qquad\qquad=[(x-1)^2-9]$
$\qquad\qquad\qquad\qquad=[(x-1)^2-(3)^2]$
$\qquad\qquad\qquad\qquad=[3^2-(x-1)^2]$
Step 2:
$\therefore I=\int \large\frac{dx}{\sqrt{3^2-(x-1)^2}}$
$\int\large\frac{dx}{\sqrt{a^2-x^2}}$$=\sin^{-1}\big(\large\frac{x}{a}\big)$$+c.$
Here $a=3,x=x-1$
$\quad\quad=\sin^{-1}\big(\large\frac{x-1}{3}\big)$$+c.$
answered Oct 3, 2013 by sreemathi.v
 
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