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Question: \[\int_{0}^{\infty}{e^{- 2x}(\sin 2x + \cos 2x)dx =}\]...

∫0∞e−2x(sin⁡2x+cos⁡2x)dx=\int_{0}^{\infty}{e^{- 2x}(\sin 2x + \cos 2x)dx =}

A

1

B

0

C

12\frac{1}{2}

D

∞\infty

Answer

12\frac{1}{2}

Explanation

Solution

∫0∞e−2x(sin⁡2x+cos⁡2x)dx\int_{0}^{\infty}{e^{- 2x}(\sin 2x + \cos 2x)dx}

=[−e−xcos⁡2x2]0∞−∫0∞(−2e−2x)(−cos⁡2x2) dx+∫0∞e−2xcos⁡2xdx=12= \left\lbrack - e^{- x}\frac{\cos 2x}{2} \right\rbrack_{0}^{\infty} - \int_{0}^{\infty}\left( - 2e^{- 2x} \right)\left( \frac{- \cos 2x}{2} \right)\ dx + \int_{0}^{\infty}{e^{- 2x}\cos 2xdx} = \frac{1}{2}.