∫1/[1+(1-x^2)^(1/2)]dx,求解答过程
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∫1/[1+(1-x^2)^(1/2)]dx,求解答过程
![∫1/[1+(1-x^2)^(1/2)]dx,求解答过程](/uploads/image/z/2755366-70-6.jpg?t=%E2%88%AB1%2F%5B1%2B%281-x%5E2%29%5E%281%2F2%29%5Ddx%2C%E6%B1%82%E8%A7%A3%E7%AD%94%E8%BF%87%E7%A8%8B)
∫1/[1+(1-x^2)^(1/2)]dx
= ∫1/[1+cost]d(sint) (令x=sint)
=∫(cost)/(1+cost)dt
=∫dt-∫dt/(1+cost)
=t+ctgt-csct+C
=arcsinx+ctgarcsinx-1/x +C
=arcsinx + (1-x^2)^(1/2)/x +C
再问: 答案是arcsinx +x/[1+ (1-x^2)^(1/2)]+C ,后面反三角函数转换不懂,可以教教我吗? =∫dt-∫dt/(1+cost) =t+ctgt-csct+C 这也不懂
再答: =∫dt-∫dt/(1+cost) =t-∫(1-cost)sintsintdt =t-∫csctcsctdt+∫d(sint)/sintsint =t+ctgt-1/sint +C =t+cost/sint-1/sint +C =arcsinx+(1-x^2)^(1/2)/x-1/x +C =arcsinx+{[(1-x^2)^(1/2)-1][(1-x^2)^(1/2)+1]}/{x[(1-x^2)^(1/2)+1]}+C =arcsinx-x/[1+ (1-x^2)^(1/2)]+C
= ∫1/[1+cost]d(sint) (令x=sint)
=∫(cost)/(1+cost)dt
=∫dt-∫dt/(1+cost)
=t+ctgt-csct+C
=arcsinx+ctgarcsinx-1/x +C
=arcsinx + (1-x^2)^(1/2)/x +C
再问: 答案是arcsinx +x/[1+ (1-x^2)^(1/2)]+C ,后面反三角函数转换不懂,可以教教我吗? =∫dt-∫dt/(1+cost) =t+ctgt-csct+C 这也不懂
再答: =∫dt-∫dt/(1+cost) =t-∫(1-cost)sintsintdt =t-∫csctcsctdt+∫d(sint)/sintsint =t+ctgt-1/sint +C =t+cost/sint-1/sint +C =arcsinx+(1-x^2)^(1/2)/x-1/x +C =arcsinx+{[(1-x^2)^(1/2)-1][(1-x^2)^(1/2)+1]}/{x[(1-x^2)^(1/2)+1]}+C =arcsinx-x/[1+ (1-x^2)^(1/2)]+C
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