A) \[\sin x-\cos x\]
B) \[\cos x-\sin x\]
C) \[-\cos x-\sin x\]
D) \[\cos x+\sin x\]
Correct Answer: A
Solution :
Given that \[\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx}=\frac{1}{2}{{e}^{x}}a+c\] ..?(i) Let \[I=\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx}=-{{e}^{x}}\cos x+\int_{{}}^{{}}{{{e}^{x}}\cos x\,dx+c}\] \[=-{{e}^{x}}\cos x+{{e}^{x}}\sin x-\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx+c}\] \[\Rightarrow 2I={{e}^{x}}(-\cos x+\sin x)+c\]. Now from (i), we get \[\frac{1}{2}{{e}^{x}}a=\frac{1}{2}{{e}^{x}}(\sin x-\cos x)\Rightarrow a=\sin x-\cos x.\]
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