第1个回答 2018-01-09
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第2个回答 2018-01-09
利用万能公式,sinx=2tan(x/2)/[1+tan²(x/2)],作换元t=tan(x/2),则x=2arctant,dx=2dt/(1+t²).当x从0变到π/2时,t从0变到1
原式=∫[0→1]2dt/(1+t²)[2+2t/(1+t²)]
=∫[0→1]dt/(1+t+t²)
=∫[0→1]dt/[(t+1/2)²+3/4]
=2/√3*arctan(2t/√3)|[0→1]
=2/√3*arctan(2/√3)
原式=∫[0→1]2dt/(1+t²)[2+2t/(1+t²)]
=∫[0→1]dt/(1+t+t²)
=∫[0→1]dt/[(t+1/2)²+3/4]
=2/√3*arctan(2t/√3)|[0→1]
=2/√3*arctan(2/√3)
第3个回答 2018-01-09
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
带入积分区间：
=(2/√3)arctan[(2/√3)tan(π/4) + 1/√3] -(2/√3)arctan[1/√3]
=(2/√3)*π/3-(2/√3)*π/6
=√3π/9本回答被提问者采纳
∫ 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
带入积分区间：
=(2/√3)arctan[(2/√3)tan(π/4) + 1/√3] -(2/√3)arctan[1/√3]
=(2/√3)*π/3-(2/√3)*π/6
=√3π/9本回答被提问者采纳
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