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प्रश्न
Integrate the following w.r.t. x : `(1)/(2sinx + sin2x)`
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उत्तर
Let I = `int (1)/(2sinx + sin2x)dx`
= `int (1)/(2sinx + 2sinx cosx)dx`
= `int (1)/(2sinx(1 + cosx))dx`
= `int (sinx)/(2sin^2x(1 + cosx))dx`
= `int (sinx.dx)/(2(1 - cos^2x)(1 + cosx))dx`
= `int (sin*dx)/(2(1 - cosx)(1 + cosx)(1 + cosx)`
= `int (sin*dx)/(2(1 - cosx)(1 + cosx)^2`
Put cos x = t
∴ – sinx .dx = dt
∴ sinx .dx = – dt
∴ I = `-(1)/(2) int (1)/((1 - t)(1 + t)^2)*dt`
= `(1)/(2) int (1)/((t - 1)(t + 1)^2)*dt`
Let `(1)/((t - 1)(t + 1)^2) = "A"/(t - 1) + "B"/(t + 1) + "C"/(t + 1)^2`
∴ 1 = A(t + 1)2 + B(t – 1)(t + 1) + C(t – 1)
Put t + 1 = 0, i.e., t = 1, we get
∴ 1 = A(0) + B(0) + C(– 2)
∴ C = `-(1)/(2)`
Put t – 1 = 0, i.e., t = 1, we get
∴ 1 = A(4) + B(0) + C(0)
∴ A = `(1)/(4)`
Comparing coefficients of t2 on both sides, we get
0 = A + B
∴ B = – A = `-(1)/(4)`
∴ `(1)/((t - 1)(t + 1)^2) = ((1/4))/(t - 1) + ((-1/4))/(t + 1) + ((-1/2))/(t + 1)^2`
∴ I = `(1)/(2) int [((1/4))/(t - 1) + ((-1/4))/(t + 1) + ((-1/2))/(t + 1)^2]*dt`
= `(1)/(8) int (1)/(t - 1)*dt - (1)/(8) int 1/(t + 1)*dt - (1)/(4) int (1)/(t - 1)^2*dt`
= `(1)/(8)log|t - 1| - (1)/(8)log|t + 1| - (1)/(4)((t + 1)^-1)/((-1)) + c`
= `(1)/(8)log|(t - 1)/(t + 1) + (1)/(4)*(1)/(t + 1) + c`
= `(1)/(8)log|(cosx - 1)/(cosx + 1)| + (1)/(4(cosx + 1)) + c`.
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