∫sin(x)^2/(x^2+1)dx from x=0 to infinity
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∫sin(x)^2/(x^2+1)dx from x=0 to infinity

[From: ] [author: ] [Date: 12-05-11] [Hit: ]
==> 0 ≤ sin^2(x)/(x^2 + 1) ≤ 1/(x^2 + 1) for all x.Since ∫(x = 0 to ∞) 1 dx/(x^2 + 1) = arctan x {for x = 0 to ∞} = π/2 (and thus convergent),the integral in question converges by the Comparison Test and is bounded above by π/2.I hope this helps!......
Decide if the integral converges or diverges. If the integral converges, find its value or give a bound on its value.

Work and explanation much appreciated, thanks!! :DD

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Note that 0 ≤ sin^2(x) ≤ 1 for all x.
==> 0 ≤ sin^2(x)/(x^2 + 1) ≤ 1/(x^2 + 1) for all x.

Since ∫(x = 0 to ∞) 1 dx/(x^2 + 1) = arctan x {for x = 0 to ∞} = π/2 (and thus convergent),
the integral in question converges by the Comparison Test and is bounded above by π/2.

I hope this helps!
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