Question

Integrate the following w.r.t. x : `(1)/(sinx + sin2x)`

Answer 1

Let I = `int (1)/(sinx + sin2x)*dx`

= `int (1)/(sinx + 2sinx cosx)*dx`

= `int dx/(sinx(1 + 2 cosx)`

= `int (sinx*dx)/(sin^2x(1 + 2cosx)`

= `int (sin*dx)/((1 - cos^2x)(1 + 2 cosx)`

= `int (sin*dx)/((1 - cosx)(1 + cosx)(1 + 2cosx)`
Put cos x = t
∴ – sinx . dx = dt
∴ sinx .dx = –  dt

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

= `-int (dt)/((1 - t)(1 + t)(1 + 2t)`

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

∴ 1 = A(1 + t)(1 + 2t) + B(1 – t)(1 + 2t) + C(1 – t)(1 + t)
Putting 1 – t = 0, i.e. t = 1, we get
1 = A(2)(3) + B(0)(3) + C(0)(2)
∴ A = `(1)/(6)`
Putting 1 – t = 0, i.e. t = – 1, we  get
1 = A(0)(– 1) + B(2)(– 1) + C(2)(0)
∴ B = `-(1)/(2)`
Putting 1 + 2t = 0, i.e. t = `-(1)/(2)`, we get

1 = `"A"(0) + "B"(0) + "C"(3/2)(1/2)`

∴ C = `(4)/(3)`

∴ `1/((1 - t)(1 + t)(1 + 2t)) = ((1/6))/(1 - t) + (((-1)/2))/(1 + t) + ((4/3))/(1 + 2t)`

∴ I = `int [((1/6))/(1 - t) + (((-1)/2))/(1 + t) + ((4/3))/(1 + 2t)]*dt`

= `-(1)/(6) int (1)/(1 - t)*dt + 1/2 int 1/(1 + t)*dt - 4/3 int 1/(1 + 2t)*dt`

= `-(1)/(6)*(log |1 - t|)/(-1) + 1/2log|1 + t| - 4/3*(log|1 + 2t|)/(2) + c`

= `-(1)/(6)log|1 - cosx| + 1/2log|1 + cosx| - 2/3log|1 + 2 cosx| + c`

= `(1)/(2)log|cosx + 1| + (1)/(6)log|cosx - 1| - (2)/(3)log|2cosx + 1| + c`.