Question

第1.2题，学霸，求解

Answer 1

（网上寻得，请展开观看）
http://www.zybang.com/question/a0c1e3c072600c3b590c72c7a313533d.html
令x = asinθ,dx = acosθdθ
原式= ∫(0→π/2) (acosθ)/(asinθ + acosθ) dθ
= (1/2)∫(0→π/2) 2cosθ/(sinθ + cosθ) dθ
= (1/2)∫(0→π/2) [(sinθ + cosθ) - (sinθ - cosθ)]/(sinθ + cosθ) dθ
= (1/2)∫(0→π/2) dθ - (1/2)∫(0→π/2) (sinθ - cosθ)/(sinθ + cosθ) dθ
= (1/2)(π/2) - (1/2)∫(0→π/2) - d(cosθ + sinθ)/(sinθ + cosθ) dθ
= π/4 + (1/2)ln(sinθ + cosθ) |(0→π/2)
= π/4 + (1/2)[ln(1 + 0) - ln(0 + 1)]
= π/4

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http://www.zybang.com/question/bf19c34255b20d9cbe69be96394b256c.html

∫[ln(1+x)/(1+x²)]dx=∫[ln(1+tanz)/(1+tan²z)]*sec²zdz (令x=tanz)
=∫ln(1+sinz/cosz)dz
=∫ln[(sinz+cosz)/cosz]dz
=∫[ln(sinz+cosz)-ln(cosz)]dz
=∫ln(sinz+cosz)dz-∫ln(cosz)dz
=∫ln[√2sin(z+π/4)]dz-∫ln(cosz)dz
=∫ln(√2)dz+∫ln[sin(z+π/4)]dz-∫ln(cosz)dz
=(π/4)ln(√2)+∫ln[sin(π/2-y)]d(-y)-∫ln(cosz)dz
(在第二个积分中,令z=π/4-y)
=πln2/8+∫ln(cosy)dy-∫ln(cosz)dz
=πln2/8+∫ln(cosz)dz-∫ln(cosz)dz
(在第一个积分中,令z=y)
=πln2/8

Answer 2

我做不来我这里没有城市的繁华和苍凉和