A) \[-\left( x+\frac{\sin 4x}{4}+\frac{\sin 2x}{2} \right)+C\]
B) \[-\left( x+\frac{\sin 4x}{4}-\frac{\sin 2x}{2} \right)+C\]
C) \[-\left( x-\frac{\sin 4x}{4}+\frac{\sin 2x}{2} \right)+C\]
D) \[-\left( x-\frac{\sin 4x}{4}-\frac{\sin 2x}{2} \right)+C\] [Note: Where 'C' is constant of integration.)
Correct Answer: A
Solution :
\[I=2\left\{ \cos (2x+\pi )+\cos \frac{2\pi }{3} \right\}\cos 2xdx\] \[=2\int_{{}}^{{}}{\left( -\cos 2x-\frac{1}{2} \right)\cos 2x\,dx}\] \[=-\int_{{}}^{{}}{\left( 2{{\cos }^{2}}2x+\cos 2x \right)}dx\] \[=-\int_{{}}^{{}}{\left( 1+\cos 4x+\cos 2x \right)}dx\] \[=-\left( x+\frac{\sin 4x}{4}+\frac{\sin 2x}{2} \right)+C\]
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