A) \[\frac{1}{2}\ln \left| \sec 2x \right|-\frac{1}{4}{{\cos }^{2}}2x+c\]
B) \[\frac{1}{2}\ln \left| \sec 2x \right|+\frac{1}{4}{{\cos }^{2}}x+c\]
C) \[\frac{1}{2}\ln \left| \cos 2x \right|-\frac{1}{4}{{\cos }^{2}}2x+c\]
D) \[\frac{1}{2}\ln \left| \cos 2x \right|+\frac{1}{4}{{\cos }^{2}}x+c\]
Correct Answer: C
Solution :
[c] \[I=\int{\frac{\cos 4x-1}{\cot x-\tan x}dx}\] \[=\int{\frac{-2{{\sin }^{2}}2x(\sin x\,\cos x)}{({{\cos }^{2}}x-{{\sin }^{2}}x)}dx}\] \[=-\int{\frac{{{\sin }^{2}}2x\sin 2x}{\cos 2x}x}\] \[=\int{\frac{({{\cos }^{2}}2x-1)\sin 2x}{\cos 2x}dx}\] Let \[t=\cos 2x\] or \[dt=-2\sin 2xdt\] \[\therefore I=\frac{1}{2}\int{\frac{(1-{{t}^{2}})}{t}}dt=\frac{1}{2}\ln \left| t \right|-\frac{{{t}^{2}}}{4}+C\] \[=\frac{1}{2}\ln \left| \cos 2x \right|-\frac{1}{4}{{\cos }^{2}}2x+c\]
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