A) \[\log \sqrt{\cos x+\sin x}+c\]
B) \[\log (\cos x-\sin x)+c\]
C) \[\log (\cos x+\sin x)+c\]
D) \[-\frac{1}{\cos x+\sin x}+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{\cos 2x}{{{(\cos x+\sin x)}^{2}}}dx}=\int_{{}}^{{}}{\frac{(\cos x-\sin x)(\cos x+\sin x)}{{{(\cos x+\sin x)}^{2}}}\text{ }}dx\] \[=\int_{{}}^{{}}{\frac{\cos x-\sin x}{\cos x+\sin x}dx}\] Put \[t=\sin x+\cos x\Rightarrow dt=(\cos x-\sin x)dx\], then it reduces to \[\int_{{}}^{{}}{\frac{1}{t}}\,\text{ }dt=\log t+c=\log (\sin x+\cos x)+c\].
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