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Question
`int (sin2x)/(3sin^4x - 4sin^2x + 1) "d"x`
Solution
Let I = `int (sin2x)/(3sin^4x - 4sin^2x + 1) "d"x`
= `int (sin 2x)/(3(sin^2x)^2 - 4sin^2x + 1) "d"x`
Put sin2x = t
∴ 2 sin x cos x dx = dt
∴ sin 2x dx = dt
∴ I = `int "dt"/(3"t"^2 - 4"t" + 1)`
= `int "dt"/(3("t"^2 - 4/3"t" + 1/3)`
`(1/2 "coefficient of " "t")^2 = [1/2 xx ((-4)/3)]^2 = 4/9`
∴ I = `1/3 int 1/("t"^2 - 4/3"t" + 4/9 - 4/9 + 1/3) "dt"`
= `1/3 int 1/(("t"^2 - 4/3"t" + 4/9) - 1/9) "dt"`
= `1/3 int 1/(("t" - 2/3)^2 - (1/3)^2) "dt"`
= `1/3*1/(2 xx 1/3) log|(("t" - 2/3) - 1/3)/(("t" - 2/3) + 1/3)| + "c"`
= `1/2 log|(3"t" - 3)/(3"t" - 1)| + "c"`
∴ I = `1/2 log|(3sin^2x - 3)/(3sin^2x - 1)| + "c"`
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