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प्रश्न
`int 1/(sinx(3 + 2cosx)) "d"x`
उत्तर
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"`
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