_ ^ 1 1 + x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{\sqrt {1 + \cos x} }}\;dx = } $

✓
$\sqrt 2 \log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
B
$\frac{1}{{\sqrt 2 }}\log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
C
$\log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
D
None of these

✓

Answer

Correct option: A.

$\sqrt 2 \log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$

a
(a)$\int_{}^{} {\frac{1}{{\sqrt {1 + \cos x} }}} \,dx = \int_{}^{} {\frac{{dx}}{{\sqrt {2{{\cos }^2}(x2)} }}} = \frac{1}{{\sqrt 2 }}\int_{}^{} {\sec \frac{x}{2}\,dx} $
$ = \frac{1}{{\sqrt 2 }}\left\{ {\log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right)} \right\}.\frac{1}{{12}} = \sqrt 2 \log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$.

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