A) \[A=-\frac{1}{2}\]
B) \[A=-\frac{1}{8}\]
C) \[A=-\frac{1}{4}\]
D) None of these
Correct Answer: B
Solution :
Given that, \[\int_{{}}^{{}}{\frac{\cos 4x+1}{\cot x-\tan x}dx=A\cos 4x+B}\]. Let\[I=\int_{{}}^{{}}{\frac{\cos 4x+1}{\cot x-\tan x}dx}\] \[=\int_{{}}^{{}}{\frac{2{{\cos }^{2}}2x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}dx=\int_{{}}^{{}}{\frac{2{{\cos }^{2}}2x}{\frac{\cos 2x}{\sin x\cos x}}dx}}\] \[=\int_{{}}^{{}}{\sin 2x\cos 2xdx}\] \[=\frac{1}{2}\int_{{}}^{{}}{\sin 4xdx}\] \[=-\frac{1}{8}\cos 4x+B\] \[\Rightarrow \]\[A\cos 4x+B=-\frac{1}{8}\cos 4x+B\][from Eq. (i)] \[\Rightarrow \]\[A=-\frac{1}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
