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