A) 1
B) 0
C) \[\frac{1}{2}\]
D) \[\infty \]
Correct Answer: C
Solution :
\[\int_{0}^{\infty }{{{e}^{-2x}}(\sin 2x+\cos 2x)dx}\] \[=\left[ -{{e}^{-x}}\frac{\cos 2x}{2} \right]_{0}^{\infty }-\int_{0}^{\infty }{\left( -2{{e}^{-2x}} \right)\,}\left( \frac{-\cos 2x}{2} \right)\text{ }dx\] \[+\int_{0}^{\infty }{{{e}^{-2x}}\cos 2x\,dx}\] \[=\frac{1}{2}\].
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