Question

高等数学定积分的一个题目，用三角万能代换公式怎么做？

Answer 1

I = ∫<0, π/2>dx/(sinx+cosx) = (1/√2)∫<0, π/2>d(x-π/4)/cos(x-π/4)
= (1/√2)[ln{sec(x-π/4)+tan(x-π/4)}]<0, π/2>
= (1/√2)[ln(√2+1)-ln(√2-1)] = √2ln(1+√2)
不必用复杂的万能公式。
一定要用万能公式， 则设 t = tan(x/2), 则 sinx = 2t/(1+t^2),
cosx = (1-t^2)/(1+t^2), dx = 2dt/(1+t^2),
I = ∫<0, π/2>dx/(sinx+cosx) = 2∫<0, 1>dt/(2t+1-t^2)
= 2∫<0, 1>dt/[2-(t-1)^2]
= (1/√2)∫<0, 1>{1/[√2-(t-1)] + 1/[√2+(t-1)]}d(t-1)
= (1/√2)[ln{√2+(t-1)]/[√2-(t-1)}]<0, 1>
= (1/√2)[0 - ln{(√2-1)/√2+1)}] = √2ln(1+√2)

Answer 2

万能公式， 则设 t = tan(x/2), 则 sinx = 2t/(1+t^2),cosx = (1-t^2)/(1+t^2), dx = 2dt/(1+t^2),I = ∫<0, π/2>dx/(sinx+cosx) = 2∫<0, 1>dt/(2t+1-t^2)= 2...

Answer 3

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Answer 4

∫(0->π/2) dx/(sinx+cosx)
=(1/√2)∫(0->π/2) dx/sin(x+π/4)
=(1/√2)∫(0->π/2) csc(x+π/4) dx
=(1/√2)[ ln|csc(x+π/4)-cot(x+π/4)| ]|(0->π/2)
=(1/√2)[ ln(√2+1) - ln(√2-1) ]
=(1/√2) ln[(√2+1)/(√2-1) ]
=(2/√2) ln(√2+1)
=√2.ln(√2+1)

Answer 5

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