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प्रश्न
Integrate the following w.r.t. x : `((3sin - 2)*cosx)/(5 - 4sin x - cos^2x)`
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उत्तर
Let I = `int ((3sin - 2)*cosx)/(5 - 4sin x - cos^2x)*dx`
= `int ((3sinx - 2)*cosx)/(5 - (1 - sin^2x) - 4sinx)*dx`
= `int ((3sinx - 2)*cosx)/(5 - 1 + sin^2x - 4sinx)*dx`
= `int ((3sinx - 2)*cosx)/(sin^2x - 4 sin x + 4)*dx`
Put sin x = t
∴ cos x dx = dt
∴ I = `int (3t - 2)/(t^2 - 4t + 4)*dt`
= `int (3t - 2)/(t - 2)^2*dt`
Let `(3t - 2)/(t - 2)^2 = "A"/(t - 2) + "B"/(t - 2)^2`
∴ 3t – 2 = A(t – 2) + B
Put t – 2 = 0, i.e. t = 2, we get
4 = A(0) + B
∴ B = 4
Put t = 0, we get
– 2 = A(– 2) + B
∴ – 2 = – 2A + 4
∴ 2A = 6
∴ A = 3
∴ `(3t - 2)/(t - 2)^2 = (3)/(t - 2) + (4)/(t - 2)^2`
∴ I = `int [3/(t - 2) + 4/(t - 2)^2]*dt`
= `3 int (1)/(t - 2)*dt + 4 int (t - 2)^-2*dt`
= `3log|t - 2| + 4*((t - 2)^-1)/(-1)*(1)/(1) + c`
= `3log|t - 2| - (4)/((t - 2)) + c`
= `3log|sin x - 2| - (4)/((sinx - 2)) + c`.
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