Question

∫(0~a)dx/{x+根号（a^2-x^2）} (a>0) 怎么算了 各位帮忙求下哈 谢谢咯

我是令 x=asint 后面的一些步骤就算不出了

Answer 1

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