Definite-Integration Question 519
Question: $ \int_0^{\infty }{{e^{-2x}}(\sin 2x+\cos 2x),dx=} $
Options:
A) 1
B) 0
C) $ \frac{1}{2} $
D) $ \infty $
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Answer:
Correct Answer: C
Solution:
$ \int_0^{\infty }{{e^{-2x}}(\sin 2x+\cos 2x)dx} $ $ =[ -{e^{-x}}\frac{\cos 2x}{2} ]_0^{\infty }-\int_0^{\infty }{( -2{e^{-2x}} ),}( \frac{-\cos 2x}{2} )dx $ $ +\int_0^{\infty }{{e^{-2x}}\cos 2x,dx} $ $ =\frac{1}{2} $ .
BETA
