_ ^ dx x - x is equal to - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{\cos x - \sin x}}} $ is equal to

✓
$\frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} + \frac{{3\pi }}{8}} \right)\,} \right| + c$
B
$\frac{1}{{\sqrt 2 }}\log \left| {\cot \left( {\frac{x}{2}} \right)\,} \right| + c$
C
$\frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} - \frac{{3\pi }}{8}} \right)\,} \right| + c$
D
$\frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} - \frac{\pi }{8}} \right)\,} \right| + c$

✓

Answer

Correct option: A.

$\frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} + \frac{{3\pi }}{8}} \right)\,} \right| + c$

a
(a) We have, $I = \int_{}^{} {\frac{{dx}}{{\cos x - \sin x}} = \frac{1}{{\sqrt 2 }}\int_{}^{} {\frac{{{d^2}}}{{\cos \left( {\frac{\pi }{4} + x} \right)}}} } $
$I = \frac{1}{{\sqrt 2 }}\int_{}^{} {\sec \left( {x + \frac{\pi }{4}} \right)} \,dx = \frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{\pi }{4} + \frac{x}{2} + \frac{\pi }{8}} \right)\,} \right| + c$
$I = \frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} + \frac{{3\pi }}{8}} \right)\,} \right| + c$.

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