Integrate the following w.r.t. x : `(1)/(sinx*(3 + 2cosx)`

#### Solution

Let I = `(1)/(sinx*(3 + 2cosx))*dx`

= `int sinx/(sin^2x*(3 + 2cosx))*dx`

= `int sinx/((1 - cos^2x)(3 + 2cosx))*dx`

= `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx`

Put cos x = t

∴ – sinx.dx = dt

∴ sinx.dx = – dt

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

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

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

∴ – 1 = A(1 + t)(3 + 2t) + B(1 - t)(3 + 2t) + C(1 - t)(1 + t)

Put 1 – t = 0, i.e. t = 1, we get

– 1 = A(2)(5) + B(0)(5) + C(0)(2)

∴ – 1 = 10A

∴ A = `(-1)/(10)`

Put 1 + t = 0, i.e. t = – 1, we get

– 1 = A(0)(1) + B(2)(1) + C(2)(0)

∴ – 1 = 2B

∴ B = `-(1)/(2)`

Put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get

– 1 = `"A"(-1/2)(0) + "B"(5/2)(0) + "C"(5/2)(-1/2)`

∴ – 1 = `-(5)/(4)"C"`

∴ C = `(4)/(5)`

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

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

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

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

= `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`.