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求∫1/2+sinx的不定积分
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求∫1/2+sinx的不定积分
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答案和解析
万能代换：令y = tan(x/2),dx = 2dy/(1 + y²),sinx = 2y/(1 + y²)
∫ 1/(2 + sinx) dx
= ∫ [2/(1 + y²)]/[2 + 2y/(1 + y²)] dy
= ∫ 1/(y² + y + 1) dy
= ∫ 1/[(y + 1/2)² + 3/4] d(y + 1/2)
= (2/√3)arctan[(y + 1/2)(2√3)] + C
= (2/√3)arctan[(2y + 1)/√3] + C
= (2/√3)arctan[(2tan(x/2) + 1)/√3] + C
= (2/√3)arctan[(2/√3)tan(x/2) + 1/√3] + C
∫ 1/(2 + sinx) dx
= ∫ [2/(1 + y²)]/[2 + 2y/(1 + y²)] dy
= ∫ 1/(y² + y + 1) dy
= ∫ 1/[(y + 1/2)² + 3/4] d(y + 1/2)
= (2/√3)arctan[(y + 1/2)(2√3)] + C
= (2/√3)arctan[(2y + 1)/√3] + C
= (2/√3)arctan[(2tan(x/2) + 1)/√3] + C
= (2/√3)arctan[(2/√3)tan(x/2) + 1/√3] + C
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