∫cost/(sint+cost)dt在0到π取积分
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∫cost/(sint+cost)dt在0到π取积分
∫ cost/(sint + cost) dt
= (1/2)∫ [(cost + sint) + (cost - sint)]/(sint + cost) dt
= (1/2)∫ [1 + (cost - sint)/(sint + cost)] dt
= t/2 + (1/2)ln|sint + cost| + C
设ƒ(t) = cost/(sint + cost)
∫(0→π) ƒ(t) dt
= ∫(0→π/2) ƒ(t) dt + ∫(π/2→π) ƒ(t) dt
= ∫(0→π/2) ƒ(t) dt + ∫(π/2→3π/4) ƒ(t) dt + ∫(3π/4→π) ƒ(t) dt
= π/4 + ∞ + ∞
= ∞
这积分发散,断续点为x = 3π/4
= (1/2)∫ [(cost + sint) + (cost - sint)]/(sint + cost) dt
= (1/2)∫ [1 + (cost - sint)/(sint + cost)] dt
= t/2 + (1/2)ln|sint + cost| + C
设ƒ(t) = cost/(sint + cost)
∫(0→π) ƒ(t) dt
= ∫(0→π/2) ƒ(t) dt + ∫(π/2→π) ƒ(t) dt
= ∫(0→π/2) ƒ(t) dt + ∫(π/2→3π/4) ƒ(t) dt + ∫(3π/4→π) ƒ(t) dt
= π/4 + ∞ + ∞
= ∞
这积分发散,断续点为x = 3π/4
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