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How solve `int_pi^(2pi)|sin n x |/x dx<=ln 2?`

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xalal | Honors

Posted June 25, 2013 at 5:45 PM via web

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How solve `int_pi^(2pi)|sin n x |/x dx<=ln 2?`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 25, 2013 at 6:03 PM (Answer #1)

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The request of the problem is vague since solving the definite integral can become a a problem because the nature of factor n is not precised.

If you need to test the inequality, you should use the following approach, such that:

`|sin nx| <= 1 => |sin nx|/x <= 1/x`

Notice that `x in [pi,2pi]` , hence, the direction of inequality is preserved when you perform the multiplication by x.

Integrating both sides yields:

`int_pi^(2pi) |sin nx|/x dx <= int_pi^(2pi) 1/x dx`

Evaluating the definite integral `int_pi^(2pi) 1/x dx` yields:

`int_pi^(2pi) 1/x dx = ln x|_pi^(2pi)`

Using the fundamental theorem of calculus, yields:

`int_pi^(2pi) 1/x dx = ln 2pi - ln pi`

Converting the difference of logarithms into a logarithm of quotient yields:

`int_pi^(2pi) 1/x dx = ln ((2pi)/(pi))`

Reducing duplicate factors yields:

`int_pi^(2pi) 1/x dx = ln 2`

Hence, replacing in inequality `ln 2` for `int_pi^(2pi) 1/x dx ` yields `int_pi^(2pi) |sin nx|/x dx <= ln 2` , thus, the given inequality holds.

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