dx x - x + 2 = - Vidyadip

MCQ

$\int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $=

A
$ - \frac{1}{{\sqrt 2 }}\tan \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$
B
$\frac{1}{{\sqrt 2 }}\tan \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$
C
$\frac{1}{{\sqrt 2 }}\cot \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$
✓
$ - \frac{1}{{\sqrt 2 }}\cot \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$

✓

Answer

Correct option: D.

$ - \frac{1}{{\sqrt 2 }}\cot \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$

(d) $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $
$ = \int {\frac{{dx}}{{\sqrt 2 (\sin x.\sin \frac{\pi }{4} - \cos x\cos \frac{\pi }{4} + 1)}}} $
$ = \frac{1}{{\sqrt 2 }}\int {\frac{{dx}}{{1 - \cos (x + \frac{\pi }{4})}}} $$ = \frac{1}{{\sqrt 2 }}\int {\frac{{dx}}{{1 - \cos 2\left( {\frac{x}{2} + \frac{\pi }{8}} \right)}}} $
$ = \frac{1}{{\sqrt 2 }}\int {\frac{{dx}}{{2{{\sin }^2}\left( {\frac{x}{2} + \frac{\pi }{8}} \right)}}} $ $ = \frac{1}{{2\sqrt 2 }}\int {{\rm{cose}}{{\rm{c}}^2}\left( {\frac{x}{2} + \frac{\pi }{8}} \right)\,dx} $
$ = \frac{1}{{2\sqrt 2 }}\frac{{ - \cot \,\left( {\frac{x}{2} + \frac{\pi }{8}} \right)\,}}{{12}} + c$$ = \frac{{ - 1}}{{\sqrt 2 }}\cot \,\left( {\frac{x}{2} + \frac{\pi }{8}} \right)\, + c$.

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