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For ∫ sin(√x) dx
let y = √x = x1/2
→ dy = 1/2 x-1/2 dx
→ 2x1/2 dy = dx
→ 2y dy = dx
→ ∫ sin(x1/2) dx = ∫(sin y) 2y dy
Now:
∫ uv dx = u∫v dx - ∫(u'∫v dx) dx
→ ∫(sin y) 2y dy = ∫2y sin y dy
= 2y ∫sin y dy - ∫(2 ∫sin y dy) dy
= -2y cos y + 2 sin y + C
= 2 sin y - 2y cos y + C
→ ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
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