A) \[\frac{1}{2}{{e}^{x}}(\sin x+\cos x)+c\]
B) \[\frac{1}{2}{{e}^{x}}(\sin x-\cos x)+c\]
C) \[{{e}^{x}}(\sin x+\cos x)+c\]
D) \[{{e}^{x}}(\sin x-\cos x)+c\]
Correct Answer: B
Solution :
Let \[I=\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx}={{e}^{x}}\sin x-\int_{{}}^{{}}{{{e}^{x}}\cos x\,dx+c}\] \[={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx+c}\] Þ \[2I={{e}^{x}}(\sin x-\cos x)+c\]Þ\[I=\frac{1}{2}{{e}^{x}}(\sin x-\cos x)+c\].
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