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