_ ^ dx x^2 - a^2 equals - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} $ equals

A
${\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + c$
✓
${\log _e}|x + \sqrt {{x^2} - {a^2}} | + c$
C
${\log _e}|x - \sqrt {{x^2} - {a^2}} | + c$
D
$\frac{{x\sqrt {{x^2} - {a^2}} }}{{2 + c}}$

✓

Answer

Correct option: B.

${\log _e}|x + \sqrt {{x^2} - {a^2}} | + c$

b
(b)$I = \int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} $.

Put$x = a\sec \theta \Rightarrow dx = a\sec \theta \,.\,\tan \theta \,d\theta $
$\therefore \,\,\,I = \int_{}^{} {\frac{{a\sec \theta \,.\,\tan \theta \,d\theta }}{{a\tan \theta }}} = \int_{}^{} {\sec \theta \,d\theta } $
$ = \log (\sec \theta + \tan \theta ) + = \log \left( {\frac{x}{a} + \frac{{\sqrt {{x^2} - {a^2}} }}{a}} \right) + c$
$ = \log (x + \sqrt {{x^2} - {a^2}} ) + c$.

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