A) \[\frac{1}{13}{{e}^{-2x}}[\sin 3x+\cos 3x]+c\]
B) \[-\frac{1}{13}{{e}^{-2x}}[\sin 3x+\cos 3x]+c\]
C) \[\frac{1}{13}{{e}^{-2x}}[2\sin 3x+3\cos 3x]+c\]
D) \[-\frac{1}{13}{{e}^{-2x}}[2\sin 3x+3\cos 3x]+c\]
Correct Answer: D
Solution :
Let \[I=\int_{{}}^{{}}{{{e}^{-2x}}\sin 3x\,dx}\] \[=-\frac{{{e}^{-2x}}\cos 3x}{3}-\int_{{}}^{{}}{\frac{2{{e}^{-2x}}\cos 3x}{3}\,dx}\] \[=-\frac{{{e}^{-2x}}\cos 3x}{3}-\frac{2}{3}\left[ \frac{{{e}^{-2x}}\sin 3x}{3}+\int_{{}}^{{}}{\frac{2{{e}^{-2x}}\sin 3x}{3}\,dx} \right]\] \[\Rightarrow I=-\frac{{{e}^{-2x}}\cos 3x}{3}-\frac{2{{e}^{-2x}}\sin 3x}{9}-\frac{4}{9}I\] \[\Rightarrow \frac{13}{9}I=-{{e}^{-2x}}\left[ \frac{3\cos 3x+2\sin 3x}{9} \right]\] Hence \[I=-\frac{1}{13}{{e}^{-2x}}[3\cos 3x+2\sin 3x]\].
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