求积分dx\/(1+根号(1-x2)) 详解过程~谢谢.
∫dx\/[1+(√1-x^2)]x=sinu √(1-x^2)=cosu tan(u\/2)=sinu\/(1+cosu)=x\/[1+√1-x^2)]=∫cosudu\/(1+cosu)=∫du-∫du\/(1+cosu)=u-∫d(u\/2)\/(cosu\/2)^2=u-tan(u\/2)+C=arcsinx-x\/[1+√(1-x^2)]+C
积分dx\/(1+根号1-x2)
原式=∫[1-√(1-x^2)]dx\/x^2 \/\/*分子分母同乘1-√(1-x^2),设x=sint,dx=costdt,(csct)^2=1\/x^2,(cott)^2=1\/x^2-1=(1-x^2)\/x^2.cott=√(1-x^2)\/x,原式=∫(1-cost)*costdt\/(sint)^2 =∫costdt\/(sint)^2-∫(cost\/sint)^2dt =∫d(sint)\/(sint)^2-...
求不定积分∫1\/(1+根号下1-x平方)
答：∫dx\/[1+√(1-x^2)] 设x=sint，-π\/2<=t<=π\/2 =∫[1\/(1+cost)]d(sint)=∫[(1+cost-1)\/(1+cost)]dt =∫[1-1\/(1+cost)]dt =t-∫(1\/cos²t\/2)d(t\/2)=t+cot(t\/2)+C =arcsinx+[1+√(1-x^2)]\/x+C ...
用换元法求不定积分 ∫ dx\/1+根号(1-X^2)
67 2017-12-26 用换元法求不定积分 ∫1\/(1-x)^dx 2013-03-03 求不定积分∫(1\/根号(1+x^2))dx 73 2013-03-31 求不定积分dx\/x根号下(x^2-1) 160 2011-11-21 求不定积分∫1\/1+根号(1-x^2)dx 190 2013-12-18 用换元法求不定积分。∫(1\/x^2 )*e^(1\/x)dx=...更多类...
求不定积分∫1\/1+根号(1-x^2)dx
∫dx\/(1+√(1-x^2))x=sinu dx=cosudu √(1-x^2)=cosu tan(u\/2)=sinu\/(1+cosu)=x\/(1+√(1-x^2))=∫cosudu\/(1+cosu)=∫[1-1\/(1+cosu)]du =u-∫du\/(1+cosu)=u-∫d(u\/2)\/(cos(u\/2))^2 =u-tan(u\/2)+C =arcsinx - x\/(1+√(1-x^2)) +C ...
1\/(1+根号里面(1-x的平方)) 的不定积分 求详细过程
dx=costdt 原式=∫[cost\/(1+cost)]dt =∫dt-∫[1\/(1+cost)]dt（上下同乘1-cost得到下面式子）=t-∫[(1-cost)\/(sin^2t)]dt =t-∫csc^2t-∫cotcsctdt =t+cott+csct+C 因为sint=x\/1,作辅助三角形，得 cott=[√(1-x^2)]\/x csct=1\/sint=1\/x 所以，原式=arcsinx+{[√...
求积分 dx\/(x+根号1-x2) 详细步骤
令x=sint dx=costdt 原式=∫costdt\/(sint+cost)令A=∫costdt\/(sint+cost) B=∫sintdt\/(sint+cost)A+B=∫(sint+cost)dt\/(sint+cost)=∫dt=t+C1 A-B=∫(cost-sint)dt\/(sint+cost)=∫d(sint+cost)\/(sint+cost)=ln|sint+cost|+C2 所以原式=A=(t+ln|sint+cost|)\/2+C ...
求积分∫1\/(x+√1-x²)dx
∫1／(x+√1-x²)dx 做三角代换：令x=sint，则√1-x²=cost，dx=costdt 原式=∫cost／(sint+cost)dt=(1\/2)∫ (cost+sint+cost-sint)／(sint+cost)dt =(1\/2)∫ (cost+sint)／(sint+cost)dt+(1\/2)∫ (cost-sint)／(sint+cost)dt =(1\/2)∫ 1dt+(1\/2)∫ 1...
dx\/(x+根号下(1-x平方))的不定积分
= ∫ 1\/[x²√(1\/x² - 1)] dx = - ∫ 1\/√[(1\/x)² - 1] d(1\/x)= - ln|1\/x + √(1\/x² - 1)| + C = ln| x\/[1 + √(1 - x²)] | + C 或设x = sinθ,dx = cosθ dθ,θ∈[- π\/2,0)U(0,π\/2]∫ 1\/[x√(1 ...
1\/(x+根号1-x^2)的积分_
S 1\/(x+根号1-x^2) dx =S cost\/(sint+cost) dt 故原式=∫cost\/(sint+cost)dt=A。令B=∫sint\/(sint+cost)dt，则A+B=∫dt=t+c1 A-B=∫(cost-sint)\/(sint+cost)dt =∫d(sint+cost)\/(sint+cost)=ln|sint+cost|+c2 所以 A=1\/2(t+ln|sint+cost|)+c 故原式=1\/2(arc...
