A) \[\frac{\sin 2x}{16}+\]constant
B) \[-\frac{\cos 2x}{4}+\]constant
C) Constant
D) None of these
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\sin 5x\cos 3x\ dx}=\frac{1}{2}\int_{{}}^{{}}{(\sin 8x+\sin 2x)}dx\] \[=\frac{-\cos 8x}{16}-\frac{\cos 2x}{4}+c\] Equating to the given value, we get \[A=\frac{-\cos 2x}{4}+c\].You need to login to perform this action.
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