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Evaluate: Int 1/"X"^2 "Sin"^2 (1/"X") "Dx" - Mathematics

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Sum

Evaluate: `int 1/"x"^2 "sin"^2 (1/"x") "dx"`

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Solution

I = `int 1/"x"^2 "sin"^2 (1/"x") "dx"`

Put `1/"x" = "t"`

`-1/"x"^2 "dx"`

∴ I = `int "sin"^2"t"  "dt" = int("cos" 2"t" - 1)/2 "dt"`

`= 1/2 ("sin"2"t")/2 - "t" + "C" = 1/4 "sin" (2/"x") - 1/"2x" + "C"`

Concept: Introduction of Integrals
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