A) \[-\frac{1}{\cos x-\sin x}+c\]
B) \[\frac{\cos x+\sin x}{\cos x-\sin x}+c\]
C) \[-\frac{1}{\sin x+\cos x}+c\]
D) \[\frac{x}{\sin x+\cos x}+c\]
E) \[\tan x+\sec x+c\]
Correct Answer: C
Solution :
Let\[I=\int{\frac{\cos x-\sin x}{{{(\cos x+\sin x)}^{2}}}}dx\] Put \[\cos x+\sin x=t\] \[\Rightarrow \] \[(\cos x-\sin x)dx=dt\] \[\therefore \] \[I=\int{\frac{1}{{{t}^{2}}}}dt\] \[I=-\frac{1}{t}+c=\frac{1}{\sin x+\cos x}+c\]
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