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
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 \[-\frac{1}{\sin x+\cos x}+c\].
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