∫0到a,1/(x+(a²-x²)½)dx(a>0),

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∫0到a,1/(x+(a²-x²)½)dx(a>0),

∫0到a,1/(x+(a²-x²)½)dx(a>0),
∫0到a,1/(x+(a²-x²)½)dx(a>0),

∫0到a,1/(x+(a²-x²)½)dx(a>0),

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令x = asiny、dx = acosy dy
∫(0→a) 1/[x + √(a² - x²)] dx
= ∫(0→π/2) (acosy dy)/(asiny + acosy)
= (1/2)∫(0→π/2) [(siny + cosy) + (- siny + cosy)]/(siny + cosy) dy
= (1/2)∫(0→π/2)...

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令x = asiny、dx = acosy dy
∫(0→a) 1/[x + √(a² - x²)] dx
= ∫(0→π/2) (acosy dy)/(asiny + acosy)
= (1/2)∫(0→π/2) [(siny + cosy) + (- siny + cosy)]/(siny + cosy) dy
= (1/2)∫(0→π/2) dy + (1/2)∫(0→π/2) d(cosy + siny)/(siny + cosy)
= (1/2)(π/2) + (1/2)ln[siny + cosy] |(0→π/2)
= π/4 + (1/2)[ln(1 + 0) - ln(0 + 1)]
= π/4
该积分的值与a无关

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设x=acost tE[0,pai] x=a时,t=0 x=0时,t=π/2
dx=-asintdt
(a^2-x^2)^1/2=asint
原式=∫(π/2,0),-asintdt/(acost+asint)
=∫(π/2,0),-sintdt/(cost+sint)
由于sint/(sint+cost)dt

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设x=acost tE[0,pai] x=a时,t=0 x=0时,t=π/2
dx=-asintdt
(a^2-x^2)^1/2=asint
原式=∫(π/2,0),-asintdt/(acost+asint)
=∫(π/2,0),-sintdt/(cost+sint)
由于sint/(sint+cost)dt
=(sint+cost-cost+sint)/[2(sint+cost)]
=-1/2[1-(cost-sint)/(sint+cost)]
不定积分=-1/2∫[1-(cost-sint)/(sint+cost)]dt
=-1/2t+1/(sint+cost)d(sint+cost)
=-1/2t+ln|sint+cost|
原式==1/2*0+ln|sin0+cos0|+1/2*pai/2-ln|1+0|
=pai/4

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