The value of _ ,0 ^ ,1 , dx x + 1 - x^2 is - Vidyadip

MCQ

The value of $\int_{\,0}^{\,1} {\,\frac{{dx}}{{x + \sqrt {1 - {x^2}} }}} $ is

A
$\frac{\pi }{3}$
B
$\frac{\pi }{2}$
C
$\frac{1}{2}$
✓
$\frac{\pi }{4}$

✓

Answer

Correct option: D.

$\frac{\pi }{4}$

d
(d) $\int_0^1 {\frac{{dx}}{{x + \sqrt {1 - {x^2}} }} = \int_0^{\pi /2} {\frac{{\cos \theta \,d\theta }}{{\sin \theta + \cos \theta }}} } $

$ = \frac{\pi }{4}$,

(Put $x = \sin \theta ,\,dx = \cos \theta \,d\theta $).

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