A) \[\frac{1}{2}\left( \cos x+\frac{1}{5}\cos 5x \right)+c\]
B) \[\frac{1}{2}\left( \cos x-\frac{1}{5}\cos 5x \right)+c\]
C) \[\cos x+\frac{1}{5}\cos 5x+c\]
D) \[\cos x-\frac{1}{5}\cos 5x+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\sin 2x\,\cos 3x\,dx}=\frac{1}{2}\int_{{}}^{{}}{2(\sin 2x\cos 3x)\,dx}\] \[=\frac{1}{2}\int_{{}}^{{}}{(\sin 5x-\sin x)\,dx}=\frac{1}{2}\left[ -\frac{\cos 5x}{5}+\cos x \right]+c\] \[=\frac{1}{2}\left[ \cos x-\frac{\cos 5x}{5} \right]+c.\]
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