Question

求解一个不定积分 x^2 / 根号下(a^2-x^2)

Answer 1

∫√（a^2-x^2）dx

设x=asint

则dx=dasint=acostdt

a^2-x^2

=a^2-a^2sint^2

=a^2cost^2

∫√（a^2-x^2）dx

=∫acost*acostdt

=a^2∫cost^2dt

=a^2∫（cos2t+1）/2dt

=a^2/4∫(cos2t+1)d2t

=a^2/4*(sin2t+2t)

将x=asint代回

∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C

扩展资料

不定积分的公式

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

Answer 2

令x=asint,(t在正负0.5pi之间） ,dx=acostdt,cost=√1-x^2/a^2
则∫x^2、√（a^2-x^2)dx
=∫a^2sin^2tacost/|a|costdt
=∫|a|sin^2t
=|a|∫(1-cos2t)/2
=|a|/2*∫(1-cos2t)
=|a|/2*[t-(sin2t)/2]
=|a|/2*[arcsinx/a-x/a*√(1-x^2/a^2)]
=|a|/2arcsinx/a-x/2√(1-x^2/a^2)