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∫sin2x3sin4x-4sin2x+1 dx

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Question

`int (sin2x)/(3sin^4x - 4sin^2x + 1)  "d"x`

Sum
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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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Chapter 2.3: Indefinite Integration - Long Answers III

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