A) \[\frac{1}{2}\log (\sin x-\cos x)+x+c\]
B) \[\frac{1}{2}[\log (\sin x-\cos x)+x]+c\]
C) \[\frac{1}{2}\log (\cos x-\sin x)+x+c\]
D) \[\frac{1}{2}[\log (\cos x-\sin x)+x]+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{\sin x\,dx}{\sin x-\cos x}}=\frac{1}{2}\int_{{}}^{{}}{\frac{2\sin x}{\sin x-\cos x}\,dx}\] \[=\frac{1}{2}\int_{{}}^{{}}{\frac{(\sin x-\cos x+\sin x+\cos x)}{\sin x-\cos x}\,dx}\] \[=\frac{1}{2}\int_{{}}^{{}}{\left( 1+\frac{\sin x+\cos x}{\sin x-\cos x} \right)\,dx}=\frac{1}{2}[x+\log (\sin x-\cos x)]+c\].
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