A) \[\frac{1}{\sin x+\cos x}+c\]
B) \[\frac{1}{\sin x-\cos x}+c\]
C) \[\log (\sin x+\cos x)+c\]
D) \[\log \left( \frac{1}{\sin x+\cos x} \right)+c\]
Correct Answer: D
Solution :
Put \[\sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt\] \[\Rightarrow \]\[-(\sin x-\cos x)dx=dt\] \[\therefore \] \[\int_{{}}^{{}}{\frac{\sin x-\cos x}{\sin x+\cos x}}\,dx=-\int_{{}}^{{}}{\frac{dt}{t}=-\log t+c=\log \left( \frac{1}{t} \right)+c}\] Hence, \[\int_{{}}^{{}}{\frac{\sin x-\cos x}{\sin x+\cos x}dx=\log \left( \frac{1}{\sin x+\cos x} \right)+c}\].
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