\[\int_{0}^{\pi /2}{{{e}^{x}}\sin x\,dx=}\] [Roorkee 1978]

\[\frac{1}{2}(1-{{e}^{\pi /2}})\]

\[\frac{1}{2}({{e}^{\pi /2}}-1)\]

\[\frac{1}{2}({{e}^{\pi /2}}+1)\]

\[\int_{0}^{\pi /2}{{{e}^{x}}\sin x\,dx=}\] �

1.	\[\int_{0}^{\pi /2}{{{e}^{x}}\sin x\,dx=}\] [Roorkee 1978]
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.	\[2({{e}^{\pi /2}}+1)\]
Answer» C. \[\frac{1}{2}(1-{{e}^{\pi /2}})\]

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\[\int_{0}^{\pi /2}{{{e}^{x}}\sin x\,dx=}\] [Roorkee 1978]

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