∫dx/2+sin2x

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∫dx/2+sin2x∫dx/2+sin2x∫dx/2+sin2x∫dx/(2+sin2x)=(1/2)∫du/(2+sinu),u=2x=(1/2)∫1/[2+2y/(1+y²)]·2/(

∫dx/2+sin2x
∫dx/2+sin2x

∫dx/2+sin2x
∫ dx/(2 + sin2x)
= (1/2)∫ du/(2 + sinu),u = 2x
= (1/2)∫ 1/[2 + 2y/(1 + y²)] · 2/(1 + y²) dy,y = tan(u/2),sinu = 2y/(1 + y²),du = 2dy/(1 + y²)
= (1/2)∫ (1 + y²)/[(1 + y²) + y] · 1/(1 + y²) dy,这是万能代换
= (1/2)∫ dy/[(y + 1/2)² + 3/4]
= (1/2) · 2/√3 · arctan[(y + 1/2)/(√3/2)] + C
= (1/√3)arctan[(2y + 1)/√3] + C
= (1/√3)arctan[(2tanx + 1)/√3] + C

万能代换