不定积分x^2dx/(a^2-x^2)^(1/2) (a>0)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/28 17:26:47
不定积分x^2dx/(a^2-x^2)^(1/2) (a>0)
![不定积分x^2dx/(a^2-x^2)^(1/2) (a>0)](/uploads/image/z/16325991-63-1.jpg?t=%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86x%5E2dx%2F%28a%5E2-x%5E2%29%5E%281%2F2%29+%EF%BC%88a%3E0%EF%BC%89)
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
= (a²/2)arcsin(x/a) - (x/2)√(a² - x²) + C
∫ 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
= (a²/2)arcsin(x/a) - (x/2)√(a² - x²) + C
不定积分dx/[x根号下(x^2+a ^2)]
不定积分 :∫ 1/(x^2-a^2)^3/2 dx
(1/(a^2-x^2)dx)的不定积分
关于不定积分的几道题求下类的不定积分1.∫dx/(x+1)(x-2)2.∫x^2dx/(a^2-x^2)^1/2(a>0
不定积分 :∫ x^2/√a^2-x^2 dx
求不定积分x^2/(a^6-x^6)dx
求不定积分∫ x/(x^2+a)dx.
∫dx/(1+√(1-2x-x²)) 不定积分 ∫tan-¹(√(a-x)/(a+x))dx 从0积
不定积分2^(1-2x)dx
不定积分dx/x(根号1-x^2)
求不定积分∫dx/(x^2+a^2)^3/2
高数求不定积分∫ 根号x^2+a^2 dx