Question

∫cost/(sint+cost)dt在0到π取积分

Answer 1

∫ 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

Answer 2

∫costdt/(sint+cost)
=∫cost(cost-sint)dt/(cost^2-sint^2)
=∫(cost^2-sintcost)dt/cos2t
=∫(1+cos2t)/2-sin2t /2dt/cos2t
=∫dt/2cos2t+t/2+(1/4)ln|cos2t|
=(1/4)ln|sec2t+tan2t|+t/2+(1/4)ln|cos2t|

∫[0,π]costdt/(sint+cost)
=(1/4)ln|sec2t+tan2t|+t/2+(1/4)ln|cos2t| |[0,π]
=(1/4)ln|sec2π|+π/2+(1/4)ln|cos2π|
-(1/4)ln|sec0|-(1/4)ln|cos0|
=π/2