A) \[-2\pi \]
B) \[\pi \]
C) \[-\pi \]
D) \[2\pi \]
Correct Answer: C
Solution :
\[I=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]}dx\] \[I=\int\limits_{0}^{\pi }{\left( \left[ \sin 2x+\sin 2x\cos 3x \right]+ \right.}\] \[\left. \left[ -\sin 2x-\sin 2x\cos 3x \right] \right)dx\]\[=\int\limits_{0}^{\pi }{-dx}=-\pi \]You need to login to perform this action.
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