A) \[\frac{1}{2}({{e}^{\pi /2}}-1)\]
B) \[\frac{1}{2}({{e}^{\pi /2}}+1)\]
C) \[\frac{1}{2}(1-{{e}^{\pi /2}})\]
D) \[2({{e}^{\pi /2}}+1)\]
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi /2}{{{e}^{x}}\sin \,x\,\,dx}\] = \[-[{{e}^{x}}\cos x]_{0}^{\pi /2}+\int_{0}^{\pi /2}{{{e}^{x}}\cos x\,dx}\] \[=-[{{e}^{x}}\cos x]_{0}^{\pi /2}+[{{e}^{x}}\sin x]_{0}^{\pi /2}-\int_{0}^{\pi /2}{{{e}^{x}}\sin x\,dx}\] \[\therefore \]\[2I=[{{e}^{x}}(\sin x-\cos x)]_{0}^{\pi /2}=({{e}^{\pi /2}}+1)\] Hence \[\int_{0}^{\pi /2}{{{e}^{x}}\sin xdx=\frac{1}{2}({{e}^{\pi /2}}+1)}\].
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