Integral Calculus Question 453
Question: $ \int_{{}}^{{}}{\frac{\cos x-\sin x}{\sqrt{\sin 2x}}\ dx} $ equals
[RPET 1996]
Options:
A) $ {{\cosh }^{-1}}(\sin x+\cos x)+c $
B) $ {{\sinh }^{-1}}(\sin x+\cos x)+c $
C) $ -{{\cosh }^{-1}}(\sin x+\cos x)+c $
D) $ -{{\sinh }^{-1}}(\sin x+\cos x)+c $
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Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\frac{\cos x-\sin x}{\sqrt{\sin 2x}}},dx $ $ =\int_{{}}^{{}}{\frac{\cos x-\sin x}{\sqrt{{{(\sin x+\cos x)}^{2}}-1}}},dx $ Put $ \sin x+\cos x=t\Rightarrow (\cos x-\sin x),dx=dt $ $ I=\int_{{}}^{{}}{\frac{dt}{\sqrt{t^{2}-1}}}={{\cosh }^{-1}}t+c={{\cosh }^{-1}}(\sin x+\cos x)+c $ .
BETA
