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DCotx等于?

C

-cotx+c

∫[cotx/(sinx)^2]dx =∫[(cosx/sinx)/(sinx)^2]dx =∫[cosx/(sinx)^3]dx =∫[1/(sinx)^3]d(sinx) =-(1/2)[1/(sinx)^2]+C =-1/[2(sinx)^2]+C =-1/(1-cos2x)+C =1/(cos2x-1)+C.

不定积分的结果应包含任意常数C ∫dx/(1+x^2)=arctanx-C1 ∫-dx/(1+x^2)=arccotx-C2 arctanx-C1+arccotx-C2=0 所以arctanx+arccotx=C1+C2 因为arctanx+arccotx=π/2 所以C1+C2=π/2 即∫dx/(1+x^2)=arctanx-C1 ∫-dx/(1+x^2)=arccotx-π/2+C1

在这里使用分部积分法即可, ∫ arctanx dx = x * arctanx - ∫ x d(arctanx) = x * arctanx - ∫ x/(1+x²) dx = x * arctanx - (1/2)∫ d(x²)/(1+x²) = x * arctanx - (1/2)∫ d(1+x²)/(1+x²) = x * arctanx - (1/2)ln(1+x...

B

解:设定积分值为w w=[0,π/2]∫2/(1+(tanx)^a)dx /**/方括号表示积分限 = [0,π/2]∫[2/(tanx)^a]/[1/(tanx)^a+1]dx = [0,π/2]∫2*(cotx)^a/[(cotx)^a+1]dx 作变量代换 u=π/2 -t ==> t= π/2 -u, 积分式化为: w= [π/2,0]∫2*[cot(π/2-u]^a/[(cot(π/2-...

d(cotx)/dx =(cotx)' =-csc²x

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