Integral Calculus Question 68
Question: $ \int{\frac{dx}{\sin x-\cos x+\sqrt{2}}} $ equals
[MP PET 2002]
Options:
A) $ -\frac{1}{\sqrt{2}}\tan ( \frac{x}{2}+\frac{\pi }{8} )+c $
B) $ \frac{1}{\sqrt{2}}\tan ( \frac{x}{2}+\frac{\pi }{8} )+c $
C) $ \frac{1}{\sqrt{2}}\cot ( \frac{x}{2}+\frac{\pi }{8} )+c $
D) $ -\frac{1}{\sqrt{2}}\cot ( \frac{x}{2}+\frac{\pi }{8} )+c $
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Answer:
Correct Answer: D
Solution:
$ 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( \frac{x}{2}+\frac{\pi }{8} )}} $
$ =\frac{1}{\sqrt{2}}\int{\frac{dx}{2{{\sin }^{2}}( \frac{x}{2}+\frac{\pi }{8} )}} $
$ =\frac{1}{2\sqrt{2}}\int{cose{c^{2}}( \frac{x}{2}+\frac{\pi }{8} ),dx} $
$ =\frac{1}{2\sqrt{2}}\frac{-\cot ,( \frac{x}{2}+\frac{\pi }{8} ),}{1/2}+c $ $ =\frac{-1}{\sqrt{2}}\cot ,( \frac{x}{2}+\frac{\pi }{8} ),+c $ .
BETA
