Question

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

Answer 1

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

sin2x = 2sinx cosx

`I = int 1/(sinx + 2sinxcosx) dx`

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

`I = intcsc x . 1/(1+2cosx) dx`

Then d(cos⁡x) = −sin⁡x dx ⇒ dx = `-dt/sin x = -csc x dt`

`I = int csc x . 1/(1+2cosx) . (-csc x) dt`

`I = -int (csc^2x)/(1+2t) dt`

But `csc^2x = 1/sin^2x = 1/(1-t^2)`

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

Partial fractions

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

Multiplying both sides by (1 − t²) (1 + 2t)

1 = A(1 + t) (1 + 2t) + B(1 − t) (1 + 2t) + C(1 − t²)

`A = 1/3, B = -1/9, C = 4/9`

`I = -int [A/(1-t) + B/(1+t) + C/(1+2t)] dt`

`I = - [-A ln |1-t| + B ln |1+t| + C/2 ln |1+2t|] + C'`

Simplify and substitute t = cos⁡x

`I = 1/3 ln |1-cosx| -1/9 ln |1+cosx| -2/9 ln |1+2cos x| + C`