急,1.∫dx/√(a²-x²)³ 2.∫x² dx/√4-x²
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急,1.∫dx/√(a²-x²)³ 2.∫x² dx/√4-x²
急,1.∫dx/√(a²-x²)³ 2.∫x² dx/√4-x²
急,1.∫dx/√(a²-x²)³ 2.∫x² dx/√4-x²
两题均是三角代换
1、设x=asinu,则√(a²-x²)³=a³cos³u,dx=acosudu
原式=∫1/(a³cos³u)*acosudu=1/a²*∫1/cos²udu=1/a²∫sec²udu=1/a²tanu+C
由sinu=x/a,知tanu=x/√(a²-x²),则原式=1/a²x/√(a²-x²)+C=x/(a²√(a²-x²))+C
2、设x=2sinu,则√(4-x²)=2cosu,dx=2cosudu
原式=∫x² dx/√4-x²=∫4sin²u/(2cosu)* 2cosudu=4∫sin²udu=2∫(1-cos2u)du
=2∫1du-∫cos2ud(2u)=2u-sin2u+C=2u-2sinucosu+C=2arcsinx-x*√(4-x²)/2+C