不定积分x^2dx/(a^2-x^2)^(1/2) （a>0）

显示全部楼层 · 2012-2-1 18:02:21

令x = a siny，dx = a cosy dy∫ x2/√(a2 - x2) dx= ∫ (a2 sin2y)(a cosy dy)/(a cosy)= a2∫ sin2y dy= (a2/2)∫ (1 - cos2y) dy= (a2/2)(y - 1/2 sin2y) + C= (a2/2)arcsin(x/a) - (a2/2)siny cosy + C= (a2/2)arcsin(x/a) - (a2/2)(x/a) √(a2 - x2)/a + C= (a&...

千问 · 2012-2-1 18:02:21

∫x^2dx/(a^2-x^2)^(1/2) （a>0）=∫[a^2/(a^2-x^2)^(1/2)]-(a^2-x^2)^(1/2)]dx=）=a^2∫1/(a^2-x^2)^(1/2)dx-∫(a^2-x^2)^(1/2)]dx=aarcsinx/a-?πa^2...