_0^2 2 + x 2 - x ,dx = - Vidyadip

MCQ

$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } \,dx = $

✓
$\pi + 2$
B
$\pi + \frac{3}{2}$
C
$\pi + 1$
D
None of these

✓

Answer

Correct option: A.

$\pi + 2$

a
(a) Put $x = 2\cos \theta $

$\Rightarrow dx = - 2\sin \theta \,d\theta ,$ then

$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } dx = - 2\int_{\pi /2}^0 {\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \sin \theta \,d\theta $

$ = 4\int_0^{\pi /2} {\frac{{\cos (\theta /2)}}{{\sin (\theta /2)}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta } $

$ = 2\int_0^{\pi /2} {(1 + \cos \theta )\,d\theta } $

$ = 2[\theta + \sin \theta ]_0^{\pi /2} $

$= 2\left[ {\frac{\pi }{2} + 1} \right] = \pi + 2$.

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