Question

Evaluate: `int ("sin 2x")/((1 + "sin x")(2 + "sin x")) "dx"`

Answer 1

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

 

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

Put sin x = t ⇒ cos x dx = dt

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

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

Put  t + 2 = 0                ⇒ t = -2

- 4 = B (-2 + 1)  ⇒ B = 4

Put t + 1 = 0                 ⇒  t = -1

-2 = A (-1 + 2) ⇒  A = -2

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

= -2 log (t + 1) + 4 log (t + 2) + C

= -2 log (sin x + 1) + 4 log (sin x + 2) + C