Question

求不定积分∫根号下(x^2-a^2) dx

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

答案：(x/2)√(x² - a²) - (a²/2)ln|x + √(x² - a²)| + C
令x = a * secz,dx = a * secztanz dz,假设x > a
∫ √(x² - a²) dx
= ∫ √(a²sec²z - a²) * (a * secztanz dz)
= a²∫ tan²z * secz dz
= a²∫ (sec²z - 1) * secz dz
= a²∫ sec³z dz - a²∫ secz dz
= a²M - a²N
M = ∫ sec³z dz = ∫ secz dtanz
= secztanz - ∫ tanz dsecz
= secztanz - ∫ tanz * (secztanz dz)
= secztanz - ∫ (sec²z - 1) * secz dz
= secztanz - M + N
2M = secztanz + N => N = (1/2)secztanz + N/2
原式= (a²/2)secztanz + a²N/2 - a²N
= (a²/2)secztanz - (a²/2)∫ secz dz
= (a²/2)secztanz - (a²/2)ln|secz + tanz| + C
= (a²/2)(x/a)[√(x² -a²)/a] - (a²/2)ln|x/a + √(x² - a²)/a| + C
= (x/2)√(x² - a²) - (a²/2)ln|x + √(x² - a²)| + C