Question:
find antiderivative of sinx divide by root 1+ sinx
Answer:
Antidarivative means integration.
So, we have to find the integration of sin x/√(1 + sin x)
=> sin x/√(1 + sin x) = ∫cos (π/2 - x)/√{1 + cos (π/2 - x)} dx
=> sin x/√(1 + sin x) = ∫ {1 + cos (π/2 - x) - 1}/√{2 * cos2 (π/2 - x)/2} dx
=> sin x/√(1 + sin x) = ∫ {2cos2 (π/4 - x/2) - 1}/√{2 * cos2 (π/4 - x/2)} dx
=> sin x/√(1 + sin x) = ∫ {2cos2 (π/4 - x/2)}/√{2 * cos2 (π/4 - x/2)} dx - ∫1/√{2 * cos2 (π/4 - x/2)} dx
=> sin x/√(1 + sin x) = ∫ √2 * cos (π/4 - x/2) dx - (1/√2)∫1/cos (π/4 - x/2)} dx
=> sin x/√(1 + sin x) = ∫ √2 * cos (π/4 - x/2) dx - (1/√2)∫sec (π/4 - x/2)} dx
=> sin x/√(1 + sin x) = √2 * sec (π/4 - x/2) - (1/√2) log |sec (π/4 - x/2) + tan(π/4 - x/2)| + C
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