A) \[k=-1/2\]
B) \[k=-1/8\]
C) \[k=-1/4\]
D) None of these
Correct Answer: B
Solution :
\[\int{\frac{1+\cos 4x}{\cot x-\tan x}}dx=\int{\frac{2{{\cos }^{2}}2x}{{{\cos }^{2}}x-{{\sin }^{2}}x}}.\sin x\cos x\,\,dx\] \[=\int{\cos 2x\sin 2x\,\,dx=-\frac{1}{8}\cos 4x+c}\], \[\therefore \]\[k=-1/8\].
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