∫(0~a)dx/{x+根号(a^2-x^2)} (a>0) 怎么算了
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/04/27 01:23:06
∫(0~a)dx/{x+根号(a^2-x^2)} (a>0) 怎么算了
我是令 x=asint 后面的一些步骤就算不出了
我是令 x=asint 后面的一些步骤就算不出了
x = a*sint,dx = a*cost dt,求微分而已
当x = 0,t = 0,当x = a,t = π/2
∫(0~a) dx/[x + √(a² - x²)]
= ∫(0~π/2) (a*cost)/(a*cost + a*sint) dt
= (1/2)∫(0~π/2) 2cost/(sint + cost) dt
= (1/2)∫(0~π/2) (cost + sint + cost - sint)/(cost + sint) dt
= (1/2)∫(0~π/2) dt + (1/2)∫(0~π/2) (cost - sint)/(cost + sint) dt
= (1/2)(π/2 - 0) + (1/2)∫(0~π/2) d(sint + cost)/(cost + sint)
= π/4 + (1/2)ln{ [cos(π/2) + sin(π/2)]/[cos(0) + sin(0)]
= π/4
当x = 0,t = 0,当x = a,t = π/2
∫(0~a) dx/[x + √(a² - x²)]
= ∫(0~π/2) (a*cost)/(a*cost + a*sint) dt
= (1/2)∫(0~π/2) 2cost/(sint + cost) dt
= (1/2)∫(0~π/2) (cost + sint + cost - sint)/(cost + sint) dt
= (1/2)∫(0~π/2) dt + (1/2)∫(0~π/2) (cost - sint)/(cost + sint) dt
= (1/2)(π/2 - 0) + (1/2)∫(0~π/2) d(sint + cost)/(cost + sint)
= π/4 + (1/2)ln{ [cos(π/2) + sin(π/2)]/[cos(0) + sin(0)]
= π/4
求定积分∫【a/0】{(根号a-根号x)^2}dx
求定积分(0-a) ∫x^2*根号下a^2-x^2 dx
求不定积分∫x^2/根号下(x^2+a^2) dx (a>0)
求定积分∫x^2[根号(a^2-x^2)]dx,上限a,下限0
定积分∫(a,0) 根号a^2-x^2 dx,a>0
1∫根号a^2-x^2dx 0到A的定积分 2 x/根号下1+x^2 dx A到0的积分
微积分,∫根号(x^2+a^2)dx
不定积分dx/[x根号下(x^2+a ^2)]
改变积分次序∫.(-a,a)dx∫(0,(根号a^2-x^2))f(x,y)dy
求∫dx/根号下(x^2+a^2),(a>0)
不定积分! ∫dx/(根号[(x-a)(b-x)]) ∫xdx/根号(5+x-x^2) ∫csc x dx
高数求不定积分∫ 根号x^2+a^2 dx