数学微分的式子积分符号下,d(x+π/2) / sin(x+π/2)=ln|csc(x+π/2)-cot(x+π/2)|
2个回答

首先令（x+π/2）=t

等价于d(t) / sin(t)=ln|csc(t)-cot(t)|+C

1/sin(t)=1/[2sin(t/2)cos(t/2)]

=cos(t/2)/[2sin(t/2)cos^2(t/2)] (同时乘以cos(t/2))

=[cos(t/2)/sin(t/2)]*[1/(2*cos^2(t/2)] (变形）

所以[1/sin(t)]*d(t)

=[cos(t/2)/sin(t/2)]*[1/(2*cos^2(t/2)]*d(t)

=[1/tan(t/2)]*[1/(2*cos^2(t/2)]*d(t)

=[1/tan(t/2)]d(tan(t/2))

积分符号下：=|lntan(t/2)|+c

而tan(t/2)=[1-cos(t)]/sin(t)=csc(t)-cot(t)

所以得证,第二个等号用三角公式就行了.

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