Question

Evaluate:

`int "dx"/(sin "x" + sin 2"x")`

Answer 1

Let, I = `int "dx"/(sin "x" + sin 2"x")`

= `int "dx"/(sin "x" + 2 sin "x" cos "x")`

= `int "dx"/(sin "x"(1 + 2 cos "x"))`

= `int (sin "x dx")/(sin^2 "x"(1 + 2 cos "x"))`

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

= `int (sin "x dx")/((1 + cos "x")(1 - cos "x")(1 + 2 cos "x"))`

Let cos x = t,

∴ − sin x dx = dt

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

Now, let

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

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

− 1 = A(1 + 3t + 2t²) + B(1 + t − 2t²) + C(1 − t²)

− 1 = (2A − 2B − C)t² + (3A + B)t + A + B + C

On comparing the coefficient of t², t and constants on both sides

2A − 2B − C = 0 ........(i)

3A + B = 0 .........(ii)

A + B + C = − 1 ..........(iii)

From equations (i), (ii) and (iii), we get

A = `-1/6`

B = `1/2`

and C = `-4/3`

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

= `-1/6 log (1 - "t") + 1/2 log (1 + "t") - 4/(3 xx 2) log (1 + 2"t") + "c"`

= `-1/6 log (1 - "t") + 1/2 log (1 + "t") - 2/3 log (1 + 2"t") + "c"`

∴ I = `-1/6 log (1 - cos "x") + 1/2 log (1 + cos "x") - 2/3 log (1 + 2 cos "x") + "c"`