∫1sinx(3+2cosx) dx - Mathematics and Statistics

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Sum

`int 1/(sinx(3 + 2cosx))  "d"x`

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Solution

Let I = `int 1/(sinx(3 + 2cosx))  "d"x`

= `int (sin x  "d"x)/(sin^2x(3 + 2cosx))`

= `int (sin x  "d"x)/((1 - cos^2x)(3 + 2cos x))`

= `int (sin x  "d"x)/((1 + cos x)(1 - cos x)(3 + 2cos x))`

Put cos x = t

∴ − sin x d x = dt 

∴ I = `int (-1)/((1 + "t")(1 - "t")(3 + 2"t"))  "dt"`

Let  `1/((1 + "t")(1 - "t")(3 + 2"t"))`

= `"A"/(1 + "t") + "B"/(1 - "t") + "C"/(3 + 2"t")`

∴ −1 = A(1 − t)(3 + 2t) + B(1 + t)(3 + 2t) + C(1 + t)(1 − t)  .......(i)

Putting t = 1 in (i), we get

−1 = 10B

∴ B = `(-1)/10`

Putting t = −1 in (i), we get

−1 = 2A

∴ A = `(-1)/2`

Putting t = `-3/2` in (i), we get

−1 = `-5/4 "C"

∴ C = `4/5`

∴ `(-1)/((1 + "t")(1 - "t")(3 + 2"t")) = ((-1)/2)/(1 + "t") + ((-1)/10)/(1 - "t") + ((-4)/5)/(3 + 2"t")`

∴ I = `int[(-1)/(2(1 + "t")) + ((-1))/(10(1 - "t")) + 4/(5(3 + 2"t"))]  "dt"`

= `-1/2 int 1/(1 + "t")  "dt" - 1/10 int 1/(1 - "t") * "dt" + 4/5 int 1/(3 + 2"t")  "dt"`

= `(-1)/2 log|1 + "t"| - 1/10 * (log|1 - "t"|)/(-1) + 4/5 * (log|3 + 2"t"|)/2 + "c"`

∴ I = `(-1)/2 log|1 + cos x| + 1/10 log|1 - cos x| + 2/5 log|3 + 2cos x| + "c"`

  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Long Answers III

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