Integral Calculus Question 201
Question: $ \int_{{}}^{{}}{\frac{\cos x-\sin x}{1+\sin 2x}\ dx=} $
[AISSE 1985]
Options:
A) $ -\frac{1}{\cos x+\sin x}+c $
B) $ \frac{1}{\cos x+\sin x}+c $
C) $ \frac{1}{\cos x-\sin x}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{\cos x-\sin x}{1+\sin 2x},dx}=\int_{{}}^{{}}{\frac{\cos x-\sin x}{{{(\sin x+\cos x)}^{2}}},dx} $ Now put $ \sin x+\cos x=t, $ then the required integral is $ -\frac{1}{\sin x+\cos x}+c $ .
BETA
