急,1.∫dx/√(a²-x²)³ 2.∫x² dx/√4-x²
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/02 21:47:38
![急,1.∫dx/√(a²-x²)³ 2.∫x² dx/√4-x²](/uploads/image/z/2590458-42-8.jpg?t=%E6%80%A5%2C1.%E2%88%ABdx%2F%E2%88%9A%28a%26%23178%3B%EF%BC%8Dx%26%23178%3B%29%26%23179%3B+2.%E2%88%ABx%26%23178%3B+dx%2F%E2%88%9A4%EF%BC%8Dx%26%23178%3B)
xRJ@ In.5%ƅzxŃx+X15i_&z7n/c{aBSlozcO^`H5#'{߬~lLf =]2}[G
oݓ,">kEGqNW(Ca6`۲5S{؊) (E$k-pyyBUx`!Uŭ@,Q,m}e3\{3bYU> [u8L̤"
B@ΤW$[eU(:7BLMⶈ|چ7W
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