A) \[-\frac{1}{16}\]
B) \[\frac{1}{16}\]
C) \[\frac{1}{8}\]
D) \[-\frac{1}{8}\]
Correct Answer: A
Solution :
Let\[I=\int_{{}}^{{}}{\frac{\cos 8x+1}{\cot 2x-\tan 2x}dx}\] Now,\[{{D}^{r}}=\cot 2x-\tan 2x=\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}\] \[=\frac{{{\cos }^{2}}2x-{{\sin }^{2}}2x}{\sin 2x\cos 2x}-\frac{2\cos 4x}{\sin 4x}\] \[\therefore \]\[I=\int_{{}}^{{}}{\frac{2{{\cos }^{2}}4x}{\frac{2\cos 4x}{\sin 4x}}dx=\int_{{}}^{{}}{\frac{2{{\cos }^{2}}4x.\sin 4x}{2\cos 4x}dx}}\] \[=\frac{1}{2}\int_{{}}^{{}}{\sin 8xdx}=-\frac{1}{2}\frac{\cos 8x}{8}+k\] \[=-\frac{1}{16}.\cos 8x+k\] \[\Rightarrow \]\[A=-\frac{1}{16}\]
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